Every worked example in the chapter, statement and solution, in the reader's own order. Cover the right-hand rule and attempt each one before you read down.
1Work by Gravity on a Projectile
A ball of mass \(m = 0.5\,\mathrm{kg}\) is thrown at \(30^\circ\) above the horizontal with speed
\(v_0 = 20\,\mathrm{m/s}\).
Find the work done by gravity as the ball rises to its maximum height.
(\(g = 10\,\mathrm{m/s^2}\).)
Key insight: gravity cares only about the vertical displacement.
The horizontal motion is irrelevant for computing \(W_g\).
Maximum height from kinematics:
\[
h = \frac{v_0^2 \sin^2\theta}{2g}
= \frac{(20)^2 \sin^2 30^\circ}{2 \times 10}
= \frac{400 \times 0.25}{20}
= 5\,\mathrm{m}
\]
Work by gravity:
\[
W_g = -mg\,\Delta h = -(0.5)(10)(5) = \boxed{-25\,\mathrm{J}}
\]
Check: negative sign is correct because the ball moves up against gravity.
Common error: using the full displacement along the parabolic arc instead of the vertical rise.
Gravity does not care about the arc length; only \(\Delta h\) matters.
2Work by a Position-Dependent Force
A force \(F(x) = (6x^2 - 4x)\,\mathrm{N}\) acts on a particle along the \(x\)-axis.
Find the work done as the particle moves from \(x = 1\,\mathrm{m}\) to \(x = 3\,\mathrm{m}\).
The force varies with position, so we integrate:
\[
W = \int_1^3 (6x^2 - 4x)\,dx
= \left[2x^3 - 2x^2\right]_1^3
\]
\[
= \bigl(2(27) - 2(9)\bigr) - \bigl(2(1) - 2(1)\bigr)
= (54 - 18) - (2 - 2) = \boxed{36\,\mathrm{J}}
\]
Graphical check:
On an \(F\)–\(x\) graph, this integral is the area under the curve \(F(x) = 6x^2 - 4x\) from \(x = 1\) to \(x = 3\).
The polynomial is positive over most of this interval, so a positive answer is expected.
Exam note:
JEE regularly gives polynomial or trigonometric \(F(x)\) and asks for work.
The computation is pure integration — the physics is in setting up the correct limits
and identifying \(F(x)\) correctly (including its sign).
3Work by Friction on a Curved Path
A \(2\,\mathrm{kg}\) block is given an initial push on a rough horizontal surface (\(\mu_k = 0.3\)).
It slides along a semicircular path of radius \(R = 1\,\mathrm{m}\) (half a circle) and comes to rest.
How much work does friction do?
(\(g = 10\,\mathrm{m/s^2}\).)
The block slides on a horizontal surface, so the normal force is \(N = mg\).
The path length of a semicircle of radius \(R\) is \(\ell = \pi R\).
\[
W_f = -\mu_k N \,\ell = -\mu_k mg \cdot \pi R
= -(0.3)(2)(10)(\pi \times 1)
= \boxed{-6\pi \approx -18.85\,\mathrm{J}}
\]
Key point: the displacement (straight-line distance from start to end of the semicircle)
is \(2R = 2\,\mathrm{m}\).
Using displacement instead of path length would give \(-12\,\mathrm{J}\) — wrong.
Friction work depends on path length, not displacement.
This is exactly the content of Perplexing Question 3.
4Block on a Rough Incline — Speed at the Bottom
A block of mass \(m = 4\,\mathrm{kg}\) starts from rest at the top of a rough incline
(length \(L = 5\,\mathrm{m}\), angle \(\theta = 30^\circ\), \(\mu_k = 0.2\)).
Find its speed at the bottom.
(\(g = 10\,\mathrm{m/s^2}\).)
Strategy: compute the work done by every force, then apply \(W_{\text{net}} = \Delta K\).
Forces and their work:
- Gravity: the height dropped is \(\Delta h = L\sin\theta = 5 \times 0.5 = 2.5\,\mathrm{m}\).
\[
W_g = mg\,\Delta h = (4)(10)(2.5) = 100\,\mathrm{J}
\]
(Positive because the block moves down: gravity aids the motion.)
- Normal force: \(W_N = 0\) (perpendicular to displacement along the incline).
- Friction: \(N = mg\cos\theta = (4)(10)\cos 30^\circ = 20\sqrt{3}\,\mathrm{N}\).
\[
W_f = -\mu_k N L = -(0.2)(20\sqrt{3})(5) = -20\sqrt{3} \approx -34.64\,\mathrm{J}
\]
Net work:
\[
W_{\text{net}} = 100 - 20\sqrt{3} \approx 65.36\,\mathrm{J}
\]
Work–Energy Theorem: the block starts from rest (\(K_i = 0\)), so
\(W_{\text{net}} = \frac{1}{2}mv_f^{\,2}\):
\[
v_f = \sqrt{\frac{2W_{\text{net}}}{m}}
= \sqrt{\frac{2(100 - 20\sqrt{3})}{4}}
= \sqrt{50 - 10\sqrt{3}}
\approx \boxed{5.72\,\mathrm{m/s}}
\]
Limiting check:
If \(\mu_k = 0\) (smooth incline): \(v_f = \sqrt{2gL\sin\theta} = \sqrt{2(10)(2.5)} = \sqrt{50} \approx 7.07\,\mathrm{m/s}\).
Our answer (\(5.72\)) is less, as expected — friction removes energy.
Compare with NLM:
The NLM approach finds \(a = g(\sin\theta - \mu_k\cos\theta) \approx 3.27\,\mathrm{m/s^2}\),
then uses \(v^2 = 2aL\) to get the same answer.
For a
straight incline, both methods are comparable in effort.
The WET becomes clearly superior when the track is curved:
gravity's work is still \(mg\Delta h\) and friction's work is still \(-f_k\ell\),
but the NLM approach would require the track's equation of curvature.
5Variable Force: Finding Final Speed
A particle of mass \(m = 2\,\mathrm{kg}\) starts from rest at \(x = 0\).
The only force acting on it along the \(x\)-axis is \(F(x) = 12\sqrt{x}\;\mathrm{N}\).
Find the speed of the particle at \(x = 4\,\mathrm{m}\).
Strategy: the force varies with position, so we cannot use \(F = ma\) directly
without solving a differential equation.
The WET bypasses this entirely.
Work done by \(F(x)\):
\[
W = \int_0^4 12\sqrt{x}\,dx = 12\int_0^4 x^{1/2}\,dx
= 12\left[\frac{2}{3}x^{3/2}\right]_0^4
= 12 \times \frac{2}{3} \times 8 = 64\,\mathrm{J}
\]
WET: since \(F(x)\) is the only force, \(W_{\text{net}} = 64\,\mathrm{J}\).
With \(K_i = 0\):
\[
v_f = \sqrt{\frac{2W}{m}} = \sqrt{\frac{2 \times 64}{2}} = \sqrt{64} = \boxed{8\,\mathrm{m/s}}
\]
Why energy wins here:
The NLM approach gives \(m\ddot{x} = 12\sqrt{x}\), a nonlinear ODE.
Solving it requires the substitution \(v\,dv = a\,dx\) — which is precisely the
derivation of the WET all over again.
The energy method skips to the answer in one integration.
Exam note:
Whenever a force depends on \(x\) and the question asks for \(v\) (not \(t\)),
the WET is almost always the fastest route.
If the question instead asks for the time to reach \(x = 4\,\mathrm{m}\),
you cannot avoid the ODE — energy alone cannot give time.
6Surprise Turning Point: Constant Force vs. Spring
A \(3\,\mathrm{kg}\) block on a smooth horizontal surface is attached to a spring (\(k = 300\,\mathrm{N/m}\))
at its natural length.
A constant horizontal force \(F = 60\,\mathrm{N}\) is applied in the direction that stretches the spring.
Find the speed of the block when the spring has stretched by \(x = 0.4\,\mathrm{m}\).
Forces doing work: the applied force \(F\) and the spring force.
(Normal and gravity are perpendicular to the horizontal motion — zero work.)
Work by applied force:
\(W_F = Fx = 60 \times 0.4 = 24\,\mathrm{J}\).
Work by spring:
\(W_s = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2 = 0 - \frac{1}{2}(300)(0.16) = -24\,\mathrm{J}\).
Net work:
\(W_{\text{net}} = 24 + (-24) = 0\).
WET: \(\Delta K = 0 \;\Rightarrow\; v_f = v_i = 0\).
This is not an error.
\(v = 0\) at \(x = 0.4\,\mathrm{m}\) means this is a turning point —
the block momentarily stops here and reverses.
The spring force has now overcome the applied force.
Where was the block fastest?
Maximum speed occurs where the net force is zero:
\[
F - kx_{\max} = 0 \qquad\Rightarrow\qquad x_{\max} = \frac{F}{k} = \frac{60}{300} = 0.2\,\mathrm{m}
\]
At \(x = 0.2\,\mathrm{m}\):
\[
W_{\text{net}} = Fx - \tfrac{1}{2}kx^2 = 60(0.2) - \tfrac{1}{2}(300)(0.04) = 12 - 6 = 6\,\mathrm{J}
\]
\[
v_{\max} = \sqrt{\frac{2 \times 6}{3}} = \sqrt{4} = \boxed{2\,\mathrm{m/s}}
\]
Lesson:
The WET gave the correct, physically meaningful answer both times.
Always trust the mathematics — and when the result surprises you, look for
the physics behind the surprise.
7Verifying Path Independence for Gravity
A \(2\,\mathrm{kg}\) block moves from \(A\) (height \(10\,\mathrm{m}\)) to \(B\) (height \(4\,\mathrm{m}\)).
Compute the work done by gravity for three paths:
(a) straight vertical drop of \(6\,\mathrm{m}\),
(b) along a \(30^\circ\) incline,
(c) a looping roller-coaster track.
(\(g = 10\,\mathrm{m/s^2}\).)
All three paths share the same start and end heights.
For each one, \(W_g = -mg\,\Delta h = -mg(h_B - h_A)\).
\[
W_g = -(2)(10)(4 - 10) = -(2)(10)(-6) = \boxed{+120\,\mathrm{J}}
\]
The answer is identical for all three paths.
This is what “conservative” means:
the work depends on the height difference \(\Delta h = -6\,\mathrm{m}\), nothing else.
Equivalently, using PE:
\(W_g = U_i - U_f = mgh_A - mgh_B = (2)(10)(10) - (2)(10)(4) = 200 - 80 = 120\,\mathrm{J}\).
Same answer, obtained by subtracting two numbers instead of integrating along a path.
This is the practical advantage of potential energy.
8Spring PE: Compression and Extension
A spring (\(k = 500\,\mathrm{N/m}\)) is compressed by \(0.1\,\mathrm{m}\) from its natural length.
The block attached to it is then released and slides on a frictionless surface.
- How much energy is stored in the spring before release?
- What is the speed of the \(0.5\,\mathrm{kg}\) block when the spring returns to its natural length?
- If the block overshoots and stretches the spring by \(0.06\,\mathrm{m}\), what is the block's speed at that point?
(a) \(U_i = \frac{1}{2}kx_i^2 = \frac{1}{2}(500)(0.01) = 2.5\,\mathrm{J}\).
(b) At natural length, \(x_f = 0\), so \(U_f = 0\).
All stored PE converts to KE (frictionless surface):
\[
\frac{1}{2}mv^2 = U_i = 2.5 \quad\Rightarrow\quad
v = \sqrt{\frac{2(2.5)}{0.5}} = \sqrt{10} \approx \boxed{3.16\,\mathrm{m/s}}
\]
(c) At \(x_f = 0.06\,\mathrm{m}\):
\(U_f = \frac{1}{2}(500)(0.0036) = 0.9\,\mathrm{J}\).
Energy conservation (frictionless):
\[
\frac{1}{2}mv^2 = U_i - U_f = 2.5 - 0.9 = 1.6\,\mathrm{J}
\quad\Rightarrow\quad
v = \sqrt{\frac{2(1.6)}{0.5}} = \sqrt{6.4} \approx \boxed{2.53\,\mathrm{m/s}}
\]
Observation:
The block is slower at \(x = 0.06\,\mathrm{m}\) than at \(x = 0\) because some KE has been
re-absorbed by the spring as elastic PE.
The total energy \(K + U\) is constant at every point —
this is the conservation law we will formalise in the next section.
9Finding Force from a Given \(U(x)\)
A particle moves along the \(x\)-axis under a conservative force whose PE is
\[
U(x) = \alpha x^2 - \beta x^4
\]
where \(\alpha = 6\,\mathrm{J/m^2}\) and \(\beta = 1\,\mathrm{J/m^4}\).
- Find the force \(F(x)\).
- Find all equilibrium positions and classify each as stable or unstable.
(a)
\[
F(x) = -\frac{dU}{dx} = -\frac{d}{dx}(\alpha x^2 - \beta x^4)
= -(2\alpha x - 4\beta x^3)
= \boxed{-2\alpha x + 4\beta x^3}
\]
(b) Equilibrium where \(F = 0\):
\[
-2\alpha x + 4\beta x^3 = 0
\qquad\Rightarrow\qquad
2x(-\alpha + 2\beta x^2) = 0
\]
Solutions: \(x = 0\) and \(x^2 = \alpha/(2\beta) = 3\), i.e. \(x = \pm\sqrt{3}\,\mathrm{m}\).
Classification via \(d^2U/dx^2\):
\[
\frac{d^2U}{dx^2} = 2\alpha - 12\beta x^2
\]
- At \(x = 0\): \(\;d^2U/dx^2 = 2\alpha = 12 \gt 0\) \(\;\Rightarrow\;\) stable equilibrium (valley).
- At \(x = \pm\sqrt{3}\): \(\;d^2U/dx^2 = 12 - 12(3) = -24 \lt 0\) \(\;\Rightarrow\;\) unstable equilibrium (hilltop).
Physical picture:
\(U(x)\) has a valley at \(x = 0\) (the particle oscillates if displaced slightly)
and two hilltops at \(x = \pm\sqrt{3}\) (the particle runs away if nudged).
We will explore this “energy landscape” analysis in depth in
another section.
10Frictionless Roller Coaster: Speed at the Lowest Point
A roller-coaster car of mass \(m = 500\,\mathrm{kg}\) starts from rest at the top of a
\(30\,\mathrm{m}\) hill.
The track is frictionless.
Find the car's speed at the bottom of the valley (ground level).
(\(g = 10\,\mathrm{m/s^2}\).)
System: car + Earth. Reference: \(U = 0\) at the valley floor.
Initial state: top of hill.
\(K_i = 0\) (starts from rest).
\(U_i = mgh = (500)(10)(30) = 150{,}000\,\mathrm{J}\).
Final state: valley floor.
\(K_f = \frac{1}{2}mv_f^{\,2}\).
\(U_f = 0\).
No friction \(\Rightarrow W_{\text{nc}} = 0\).
Conservation:
\[
0 + 150{,}000 = \tfrac{1}{2}(500)\,v_f^{\,2} + 0
\]
\[
v_f = \sqrt{\frac{2 \times 150{,}000}{500}} = \sqrt{600} \approx \boxed{24.5\,\mathrm{m/s}}
\]
Notice: the mass cancels. \(v_f = \sqrt{2gh}\) regardless of \(m\).
A \(50\,\mathrm{kg}\) child and a \(5000\,\mathrm{kg}\) truck released from the same height
reach the bottom at the same speed.
This is the energy-method cousin of Galileo's observation that all objects
fall with the same acceleration.
11Block + Spring + Rough Surface
A \(2\,\mathrm{kg}\) block compresses a horizontal spring (\(k = 800\,\mathrm{N/m}\))
by \(x_0 = 0.25\,\mathrm{m}\) and is released from rest.
The first \(0.25\,\mathrm{m}\) of floor (under the spring) is smooth.
Beyond that, the floor is rough with \(\mu_k = 0.4\).
How far past the spring's natural length does the block slide before stopping?
(\(g = 10\,\mathrm{m/s^2}\).)
System: block + spring + Earth.
Reference: \(U_{\text{spring}} = 0\) at natural length; \(U_{\text{gravity}} = 0\)
at floor level (no height change, so \(\Delta U_g = 0\) throughout).
Initial state: spring compressed by \(x_0\), block at rest.
\[
K_i = 0, \quad U_i = \tfrac{1}{2}kx_0^2 = \tfrac{1}{2}(800)(0.0625) = 25\,\mathrm{J}
\]
Final state: block at rest on the rough floor, spring at natural length (\(U_f = 0\)).
\[
K_f = 0, \quad U_f = 0
\]
Non-conservative work:
The rough patch has length \(d\) (the unknown).
Friction acts only on this patch:
\(W_f = -\mu_k mg\,d = -(0.4)(2)(10)\,d = -8d\).
Generalised energy equation:
\[
K_i + U_i + W_f = K_f + U_f
\]
\[
0 + 25 - 8d = 0 + 0
\]
\[
d = \frac{25}{8} = \boxed{3.125\,\mathrm{m}}
\]
Limiting checks:
- If \(\mu_k = 0\): \(d \to \infty\). Correct — on a frictionless floor the block never stops (it oscillates).
- If \(k \to \infty\) (very stiff spring): for the same compression \(x_0\),
\(U_i = \frac{1}{2}kx_0^2 \to \infty\) and \(d \to \infty\). Correct — more stored energy means more sliding.
- If \(x_0 \to 0\): \(d \to 0\). Correct — no compression, no motion.
12Pendulum: Speed at the Bottom
A simple pendulum of length \(L = 1\,\mathrm{m}\) and bob mass \(m = 0.2\,\mathrm{kg}\)
is released from rest at an angle \(\theta_0 = 60^\circ\) from the vertical.
Find the speed of the bob at the lowest point.
(\(g = 10\,\mathrm{m/s^2}\).)
System: bob + Earth. Reference: \(U = 0\) at the lowest point.
Geometry:
When the string makes angle \(\theta_0\) with the vertical, the bob is at height
\(h = L - L\cos\theta_0 = L(1 - \cos\theta_0)\) above the lowest point.
\[
h = 1(1 - \cos 60^\circ) = 1 - 0.5 = 0.5\,\mathrm{m}
\]
Initial: \(K_i = 0\), \(U_i = mgh = (0.2)(10)(0.5) = 1\,\mathrm{J}\).
Final (lowest point): \(K_f = \frac{1}{2}mv^2\), \(U_f = 0\).
Forces: gravity (conservative) and tension.
Tension is always perpendicular to the bob's velocity (radial vs. tangential),
so \(W_T = 0\). Hence \(W_{\text{nc}} = 0\).
\[
0 + 1 = \tfrac{1}{2}(0.2)\,v^2 + 0
\quad\Rightarrow\quad
v = \sqrt{\frac{2}{0.2}} = \sqrt{10} \approx \boxed{3.16\,\mathrm{m/s}}
\]
General formula:
\(v = \sqrt{2gL(1 - \cos\theta_0)}\).
Mass cancels — a heavier bob swings at the same speed (same \(g\), same geometry).
What energy cannot tell you:
The speed at the bottom is \(\sqrt{10}\,\mathrm{m/s}\).
But what is the tension in the string there?
For that, you need NLM at the lowest point:
\[
T - mg = \frac{mv^2}{L}
\quad\Rightarrow\quad
T = mg + \frac{mv^2}{L} = (0.2)(10) + \frac{(0.2)(10)}{1} = 2 + 2 = 4\,\mathrm{N}
\]
This is the combo: energy for speed, then NLM for force.
We will devote an entire section to this technique later (another section).
13Rough Curved Track: Energy Lost to Friction
A \(4\,\mathrm{kg}\) block starts from rest at the top of a curved track (\(h = 5\,\mathrm{m}\))
and slides to the bottom.
The track length (arc length) is \(\ell = 8\,\mathrm{m}\), and \(\mu_k = 0.15\) throughout.
Find the speed at the bottom.
(\(g = 10\,\mathrm{m/s^2}\).)
Reference: \(U = 0\) at the bottom.
\(K_i = 0\), \(U_i = mgh = (4)(10)(5) = 200\,\mathrm{J}\).
\(K_f = \frac{1}{2}(4)v^2\), \(U_f = 0\).
Normal force on a curved track varies, but for a gentle curve
\(N \approx mg\cos\alpha\) at each local angle \(\alpha\).
As a simplification standard in JEE problems,
when \(\mu_k\) and the arc length are given,
\(W_f = -\mu_k mg\,\ell\) is used with \(N \approx mg\)
(valid when the track curvature is gentle, i.e. centripetal corrections are small).
\[
W_f = -(0.15)(4)(10)(8) = -48\,\mathrm{J}
\]
\[
0 + 200 - 48 = 2v^2 + 0
\quad\Rightarrow\quad
v = \sqrt{\frac{152}{2}} = \sqrt{76} \approx \boxed{8.72\,\mathrm{m/s}}
\]
Compare with frictionless case:
\(v_{\text{smooth}} = \sqrt{2gh} = \sqrt{100} = 10\,\mathrm{m/s}\).
Friction removes \(48\,\mathrm{J}\) out of \(200\,\mathrm{J}\) — a \(24\%\) energy loss.
Why energy shines here:
The track is curved, so the normal force changes at every point.
An NLM approach would require knowing the track equation and integrating
\(\mu_k N(\alpha)\,ds\) along the arc — a nightmare.
The energy method bypasses all of this with a single friction-work term.
14Engine Power and Maximum Speed
A car of mass \(1200\,\mathrm{kg}\) has a maximum engine power of \(90\,\mathrm{kW}\).
The total resistive force at speed \(v\) is \(f = 0.5\,v^2\;\mathrm{N}\).
Find the maximum speed on a level road.
At maximum speed the car is in equilibrium: engine force \(=\) drag.
All engine power goes into overcoming drag:
\[
P_{\max} = f \times v_{\max} = 0.5\,v_{\max}^2 \times v_{\max} = 0.5\,v_{\max}^3
\]
\[
v_{\max}^3 = \frac{P_{\max}}{0.5} = \frac{90{,}000}{0.5} = 180{,}000
\]
\[
v_{\max} = (180{,}000)^{1/3} \approx \boxed{56.5\,\mathrm{m/s} \approx 203\,\mathrm{km/h}}
\]
Check: at \(v_{\max}\), \(f = 0.5(56.5)^2 \approx 1594\,\mathrm{N}\).
\(P = fv = 1594 \times 56.5 \approx 90\,\mathrm{kW}\). Consistent.
15Pump Lifting Water
A pump lifts water from a well \(20\,\mathrm{m}\) deep at a rate of \(10\,\mathrm{kg/s}\)
and delivers it at ground level with a speed of \(4\,\mathrm{m/s}\).
Find the minimum power of the pump.
(\(g = 10\,\mathrm{m/s^2}\).)
Each second, the pump must provide:
PE increase: \(\Delta U = \dot{m}\,g\,h = (10)(10)(20) = 2000\,\mathrm{W}\).
KE delivered: \(\Delta K = \frac{1}{2}\dot{m}\,v^2 = \frac{1}{2}(10)(16) = 80\,\mathrm{W}\).
Minimum power:
\[
P_{\min} = \Delta U + \Delta K = 2000 + 80 = \boxed{2080\,\mathrm{W} \approx 2.08\,\mathrm{kW}}
\]
Note: “minimum” power assumes no friction or turbulence losses.
A real pump would need more.
The PE term dominates — kinetic energy at the outlet is a small correction.
16Complete \(U(x)\) Analysis
A particle of mass \(m\) moves along the \(x\)-axis under a conservative force
whose potential energy is
\[
U(x) = U_0\left[\left(\frac{x}{a}\right)^2 - \left(\frac{x}{a}\right)^4\right]
\]
where \(U_0 \gt 0\) and \(a \gt 0\) are constants.
- Find all equilibrium positions.
- Classify each as stable or unstable.
- If the particle has total energy \(E = U_0/4\), find the turning points.
- Describe the motion qualitatively.
Let \(\xi = x/a\) for convenience, so \(U = U_0(\xi^2 - \xi^4)\).
(a) Equilibria: set \(dU/dx = 0\).
\[
\frac{dU}{dx} = \frac{U_0}{a}(2\xi - 4\xi^3) = \frac{2U_0}{a}\,\xi(1 - 2\xi^2) = 0
\]
Solutions: \(\xi = 0\) and \(\xi^2 = 1/2\), i.e. \(x = 0\) and \(x = \pm a/\sqrt{2}\).
(b) Classification:
\[
\frac{d^2U}{dx^2} = \frac{U_0}{a^2}(2 - 12\xi^2)
\]
- At \(x = 0\) (\(\xi = 0\)): \(d^2U/dx^2 = 2U_0/a^2 \gt 0\) \(\Rightarrow\) stable (valley).
- At \(x = \pm a/\sqrt{2}\) (\(\xi^2 = 1/2\)): \(d^2U/dx^2 = U_0(2 - 6)/a^2 = -4U_0/a^2 \lt 0\)
\(\Rightarrow\) unstable (hilltops).
The hilltop value: \(U(\pm a/\sqrt{2}) = U_0(1/2 - 1/4) = U_0/4\).
(c) Turning points for \(E = U_0/4\):
Set \(U(x) = E\):
\[
U_0(\xi^2 - \xi^4) = U_0/4
\quad\Rightarrow\quad
\xi^4 - \xi^2 + 1/4 = 0
\quad\Rightarrow\quad
(\xi^2 - 1/2)^2 = 0
\]
So \(\xi^2 = 1/2\), i.e. \(x = \pm a/\sqrt{2}\).
These are the hilltop positions themselves.
The turning “points” coincide with the unstable equilibria.
(d) Qualitative motion:
With \(E = U_0/4\), the particle is confined to \(|x| \le a/\sqrt{2}\).
It oscillates in the valley about \(x = 0\),
turning around at \(x = \pm a/\sqrt{2}\) where \(K = 0\).
The motion is bound.
If \(E\) were slightly larger than \(U_0/4\), the particle would have enough energy
to cross the hilltops and escape — the motion would become unbound.
\(E = U_0/4\) is the critical energy separating bound from unbound orbits.
17Force from \(U(x)\) and Motion Prediction
A particle has \(U(x) = 5x^2 - x^3\) (SI units).
- Find the force.
- Find the equilibrium positions and classify them.
- If the particle is at \(x = 1\,\mathrm{m}\) with \(K = 0\), in which direction does it move?
(a) \(F(x) = -dU/dx = -(10x - 3x^2) = 3x^2 - 10x\).
(b) Set \(F = 0\): \(x(3x - 10) = 0\), so \(x = 0\) and \(x = 10/3\,\mathrm{m}\).
\(d^2U/dx^2 = 10 - 6x\).
- At \(x = 0\): \(10 \gt 0\) \(\Rightarrow\) stable.
- At \(x = 10/3\): \(10 - 20 = -10 \lt 0\) \(\Rightarrow\) unstable.
(c) At \(x = 1\): \(F(1) = 3 - 10 = -7\,\mathrm{N}\).
The force is negative, so the particle moves in the \(-x\) direction
(toward the stable equilibrium at \(x = 0\)).
18Spring Bomb on a Smooth Table
Two blocks of masses \(m_1 = 2\,\mathrm{kg}\) and \(m_2 = 3\,\mathrm{kg}\) are held together on a smooth horizontal surface with a compressed spring (\(k = 500\,\mathrm{N/m}\), compression \(x_0 = 0.1\,\mathrm{m}\)) between them.
When released, find the speed of each block after separation.
System: both blocks + spring.
External work: The normal force and gravity are perpendicular to the motion; no external force does work. Hence \(W_{\text{ext}} = 0\).
Internal work: The spring force is conservative. No friction. So \(W_{\text{int, nc}} = 0\).
Therefore \(\Delta K + \Delta U = 0\):
\[
\bigl(\tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2 \bigr) - 0 = \tfrac{1}{2}k x_0^2 - 0.
\]
\[
\tfrac{1}{2}(2)v_1^2 + \tfrac{1}{2}(3)v_2^2 = \tfrac{1}{2}(500)(0.01) = 2.5\,\mathrm{J}.
\]
One equation, two unknowns—energy alone is insufficient.
We need momentum conservation (no external horizontal force):
\[
m_1 v_1 = m_2 v_2 \quad\Longrightarrow\quad 2v_1 = 3v_2 \quad\Longrightarrow\quad v_1 = \tfrac{3}{2}v_2.
\]
Substituting (ii) into (i):
\[
(2)\!\left(\tfrac{3}{2}v_2\right)^2 \!/2 + (3)v_2^2/2 = 2.5
\quad\Longrightarrow\quad
\tfrac{9}{4}v_2^2 + \tfrac{3}{2}v_2^2 = 2.5
\quad\Longrightarrow\quad
\tfrac{15}{4}v_2^2 = 2.5.
\]
\[
v_2 = \sqrt{\frac{10}{15}} = \sqrt{\frac{2}{3}} \approx 0.816\,\mathrm{m/s},
\qquad
v_1 = \tfrac{3}{2}v_2 = \sqrt{\frac{3}{2}} \approx 1.225\,\mathrm{m/s}.
\]
Exam insight: Energy conservation alone cannot solve a two-body spring release—you always need a second equation. That second equation is momentum conservation, which holds because there is no net external force in the direction of motion. Every JEE/NEET problem on spring bombs tests whether you recognise this two-equation structure.
19Block–Spring Collision
A block of mass \(m = 1\,\mathrm{kg}\) slides at \(v_0 = 4\,\mathrm{m/s}\) on a smooth surface and strikes a spring (\(k = 200\,\mathrm{N/m}\)) attached to a wall.
Find the maximum compression of the spring and the speed of the block when the spring is compressed by half the maximum amount.
System: block + spring + wall (wall is fixed, so it does no work).
At maximum compression \(x_{\max}\), the block is momentarily at rest:
\[
\tfrac{1}{2}mv_0^2 = \tfrac{1}{2}k x_{\max}^2
\quad\Longrightarrow\quad
x_{\max} = v_0\sqrt{\frac{m}{k}} = 4\sqrt{\frac{1}{200}} = \frac{4}{10\sqrt{2}} = 0.2\sqrt{2} \approx 0.283\,\mathrm{m}.
\]
At half maximum compression (\(x = x_{\max}/2\)):
\[
\tfrac{1}{2}mv_0^2 = \tfrac{1}{2}mv^2 + \tfrac{1}{2}k\!\left(\frac{x_{\max}}{2}\right)^2.
\]
\[
\tfrac{1}{2}mv^2 = \tfrac{1}{2}mv_0^2 - \tfrac{1}{2}k\cdot\frac{x_{\max}^2}{4}
= \tfrac{1}{2}mv_0^2 - \frac{1}{4}\cdot\tfrac{1}{2}kx_{\max}^2
= \tfrac{1}{2}mv_0^2\!\left(1 - \frac{1}{4}\right)
= \frac{3}{4}\cdot\tfrac{1}{2}mv_0^2.
\]
\[
v = v_0\sqrt{\frac{3}{4}} = 4\cdot\frac{\sqrt{3}}{2} = 2\sqrt{3} \approx 3.46\,\mathrm{m/s}.
\]
Physical insight: At half the maximum compression, three-quarters of the initial kinetic energy remains kinetic. The spring has absorbed only one-quarter. Energy storage is quadratic in compression—the first half of compression stores only \(\frac{1}{4}\) of the total energy, while the second half stores \(\frac{3}{4}\). This nonlinearity is a frequent source of wrong intuition.
20Atwood Machine via Energy
Two masses \(m_1 \gt m_2\) are connected by a light, inextensible string over a smooth, massless pulley.
Starting from rest, find the speed of each mass after the heavier mass has descended a height \(h\).
System: \(m_1 + m_2\) (include the Earth for gravitational PE).
Since the string is inextensible, both masses move with the same speed \(v\) at every instant.
The string tension is internal to the system and does zero net work: it does \(+Th\) on one mass and \(-Th\) on the other. (This is the payoff of system thinking—tension drops out entirely.)
No friction, no external work. Conservation of mechanical energy:
\[
\Delta K + \Delta U = 0.
\]
\[
\bigl(\tfrac{1}{2}m_1 v^2 + \tfrac{1}{2}m_2 v^2\bigr) + \bigl(-m_1 g h + m_2 g h\bigr) = 0.
\]
\[
\tfrac{1}{2}(m_1 + m_2)v^2 = (m_1 - m_2)gh.
\]
\[
\boxed{v = \sqrt{\frac{2(m_1 - m_2)gh}{m_1 + m_2}}}
\]
Sanity checks:
(i) If \(m_1 = m_2\): \(v = 0\)—balanced masses, no motion. ✓
(ii) If \(m_2 = 0\): \(v = \sqrt{2gh}\)—free fall. ✓
(iii) If \(m_1 \gg m_2\): \(v \to \sqrt{2gh}\)—the light mass is negligible. ✓
Exam note: The energy method gives the speed in one line. The force method (free-body diagrams, \(m_1 a = m_1 g - T\), \(m_2 a = T - m_2 g\), solve for \(a\), then use \(v^2 = 2ah\)) takes three. Know both; use energy when only speeds matter.
21Block on Table, Mass Hanging — Energy Method
A block of mass \(M = 4\,\mathrm{kg}\) rests on a smooth horizontal table.
It is connected by a string over a frictionless pulley at the edge to a hanging mass \(m = 1\,\mathrm{kg}\).
Find the speed of the system when the hanging mass has fallen \(2\,\mathrm{m}\) from rest.
(\(g = 10\,\mathrm{m/s^2}\).)
System: \(M + m + \text{Earth}\).
Both bodies share the same speed \(v\) (inextensible string).
The block on the table has no change in gravitational PE.
Only the hanging mass loses height \(h = 2\,\mathrm{m}\).
\[
0 + mgh = \tfrac{1}{2}(M + m)v^2.
\]
\[
v = \sqrt{\frac{2mgh}{M+m}} = \sqrt{\frac{2(1)(10)(2)}{5}} = \sqrt{8} = 2\sqrt{2} \approx 2.83\,\mathrm{m/s}.
\]
Where did the PE go?
It became KE of both bodies, not just the hanging mass.
The string redistributes the energy: the internal tension does positive work on the table block and negative work on the hanging mass, transferring kinetic energy from one to the other.
But the total mechanical energy is conserved, so the transfer is invisible at the system level.
22Rough Block on a Rough Block
A small block of mass \(m = 2\,\mathrm{kg}\) sits on top of a large block of mass \(M = 8\,\mathrm{kg}\), which rests on a smooth floor.
Between the two blocks \(\mu_s = 0.4\) and \(\mu_k = 0.3\); the floor is frictionless.
A horizontal force \(F\) is applied to the lower block. Take \(g = 10\,\mathrm{m/s^2}\).
(a) With \(F = 20\,\mathrm{N}\): do the surfaces slip? Find the speed of each block and the heat generated after the lower block has moved \(1\,\mathrm{m}\).
(b) Above what \(F\) do they slip?
(c) Repeat (a) for \(F = 60\,\mathrm{N}\).
System: both blocks. Friction between them is internal; the floor is smooth, so the only external force doing work is \(F\) on \(M\).
Step 0 — is the contact sliding or not?
You cannot write \(f = \mu_k N\) until you know the surfaces are sliding, and \(\mu_k\) cannot tell you that. Assume they move together and test whether static friction can deliver it: \(a = F/(M+m)\) and \(f_{\text{needed}} = ma = mF/(M+m)\), friction being the only horizontal force on the upper block. The contact can supply at most \(f_{\max} = \mu_s m g = 8\,\mathrm{N}\).
(a) \(F = 20\,\mathrm{N}\). \(a = 2\,\mathrm{m/s^2}\), so \(f_{\text{needed}} = 4\,\mathrm{N} \lt 8\,\mathrm{N}\): they move together, \(d_{\text{rel}} = 0\), no heat. The friction force is \(4\,\mathrm{N}\) — what the motion demands — not \(\mu_k m g\).
\(v = \sqrt{2as} = 2\,\mathrm{m/s}\) for both. \(\Delta K = \frac{1}{2}(10)(4) = 20\,\mathrm{J} = W_{\text{ext}}\), so the audit closes with \(Q = 0\). ✓
(b) Threshold. \(mF/(M+m) = \mu_s m g \implies F_{\text{slip}} = \mu_s g(M+m) = 40\,\mathrm{N}\). \(m\) cancels: it depends on the total mass, not the carried mass.
(c) \(F = 60\,\mathrm{N}\). Now \(f = \mu_k m g = 6\,\mathrm{N}\) is the right model. \(M\) is the block being pushed, so \(m\) lags and friction on it acts forward: \(a_m = 3\), \(a_M = (60-6)/8 = 6.75\,\mathrm{m/s^2}\). \(a_M \gt a_m\) confirms the assumed sliding direction.
At \(s_M = 1\,\mathrm{m}\): \(t^2 = 8/27\), \(v_M = \sqrt{13.5} \approx 3.67\,\mathrm{m/s}\), \(v_m = \sqrt{8/3} \approx 1.63\,\mathrm{m/s}\), \(s_m = 4/9\,\mathrm{m}\).
\(d_{\text{rel}} = 1 - 4/9 = 5/9\,\mathrm{m}\) — the difference, never either displacement alone — so \(Q = 6 \times 5/9 = 10/3 \approx 3.33\,\mathrm{J}\).
\(\Delta K = 54 + 8/3 = 170/3\,\mathrm{J}\), and \(\Delta K + Q = 60\,\mathrm{J} = W_{\text{ext}}\). ✓
Lesson: the friction model is a consequence of the contact regime, so decide the regime first — from \(\mu_s\), never from \(\mu_k\). Two errors follow from the wrong order: \(\mu_k m g\) on a contact that is not sliding (part (a): \(6\,\mathrm{N}\) instead of \(4\,\mathrm{N}\), and heat that never existed), and \(Q\) from one block's displacement rather than the relative one (part (c): \(6\,\mathrm{J}\) instead of \(3.33\,\mathrm{J}\), after which the audit refuses to close).
NEET / AP Physics C Style
23Work by a force at an angle
A \(4\,\mathrm{kg}\) crate is pulled \(6\,\mathrm{m}\) along a level floor by a \(25\,\mathrm{N}\) force directed \(37^\circ\) above the horizontal. The floor has \(\mu_k = 0.20\). Find the work done by (a) the applied force, (b) friction, and (c) the net work. (\(\sin37^\circ = 0.6,\ \cos37^\circ = 0.8\).)
Given: \(m = 4\,\mathrm{kg}\), \(d = 6\,\mathrm{m}\), \(F = 25\,\mathrm{N}\) at \(37^\circ\), \(\mu_k = 0.20\).
Find: \(W_{\text{app}}\), \(W_{\text{fric}}\), \(W_{\text{net}}\).
Setup: Only the horizontal component of \(F\) moves the crate, so \(W_{\text{app}} = Fd\cos\theta\). The upward component of \(F\) lightens the normal force, which sets the friction. Gravity and the normal force are perpendicular to the (horizontal) displacement and do no work.
Solve:
\[ W_{\text{app}} = Fd\cos\theta = 25 \times 6 \times 0.8 = 120\,\mathrm{J}. \]
Normal force: \(N = mg - F\sin\theta = 40 - 25(0.6) = 25\,\mathrm{N}\), so \(f = \mu_k N = 0.20 \times 25 = 5\,\mathrm{N}\) and
\[ W_{\text{fric}} = -fd = -5 \times 6 = -30\,\mathrm{J}. \]
Answer: \(\boxed{W_{\text{app}} = 120\,\mathrm{J},\quad W_{\text{fric}} = -30\,\mathrm{J},\quad W_{\text{net}} = 90\,\mathrm{J}}\)
Check: Dimensions: \(\mathrm{N\cdot m = J}\) ✓. Sign sanity: the horizontal pull \(F\cos\theta = 20\,\mathrm{N}\) exceeds friction \(5\,\mathrm{N}\), so the net work is positive and the crate speeds up, as found. ✓
24Work–energy theorem: speed from net work
A \(2\,\mathrm{kg}\) block, starting from rest, is pushed \(4\,\mathrm{m}\) across a floor by a constant horizontal force of \(12\,\mathrm{N}\) against a friction force of \(4\,\mathrm{N}\). Find its final speed.
Given: \(m = 2\,\mathrm{kg}\), \(u = 0\), \(d = 4\,\mathrm{m}\), \(F = 12\,\mathrm{N}\), \(f = 4\,\mathrm{N}\).
Find: the final speed \(v\).
Setup: The net work equals the change in kinetic energy; no need to find the acceleration first.
Solve:
\[ W_{\text{net}} = (F - f)d = (12-4)(4) = 32\,\mathrm{J} = \tfrac{1}{2}mv^2 \;\Rightarrow\; v = \sqrt{\frac{2(32)}{2}} = \sqrt{32}\,\mathrm{m/s}. \]
Answer: \(\boxed{v = 4\sqrt{2} \approx 5.7\,\mathrm{m/s}}\)
Check: Alternative method: \(a = (F-f)/m = 8/2 = 4\,\mathrm{m/s^2}\), so \(v^2 = 2ad = 2(4)(4) = 32\), giving the same \(v = \sqrt{32}\). ✓
25Energy conservation on a frictionless wire
A bead slides from rest down a frictionless wire starting at height \(1.8\,\mathrm{m}\) above the ground. Find its speed (a) at the ground and (b) at the point where the wire is \(1.2\,\mathrm{m}\) above the ground.
Given: frictionless, \(h_0 = 1.8\,\mathrm{m}\), starts from rest.
Find: speed at \(h = 0\) and at \(h = 1.2\,\mathrm{m}\).
Setup: With only gravity doing work, \(\tfrac12 mv^2 = mg\,\Delta h\); the mass cancels, so the bead's mass is irrelevant.
Solve:
\[ v_{\text{ground}} = \sqrt{2g h_0} = \sqrt{2(10)(1.8)} = \sqrt{36} = 6\,\mathrm{m/s}, \]
\[ v_{1.2} = \sqrt{2g(h_0 - h)} = \sqrt{2(10)(0.6)} = \sqrt{12} = 2\sqrt{3}\,\mathrm{m/s}. \]
Answer: \(\boxed{v_{\text{ground}} = 6\,\mathrm{m/s},\quad v_{1.2} = 2\sqrt{3} \approx 3.5\,\mathrm{m/s}}\)
Check: Energy bookkeeping at \(1.2\,\mathrm{m}\) (per unit mass): \(\tfrac12(12) + g(1.2) = 6 + 12 = 18 = g h_0 = 10(1.8)\) ✓. The bead is slower higher up, as expected. ✓
26Spring launch on a frictionless table
A spring of stiffness \(k = 200\,\mathrm{N/m}\) is compressed \(0.10\,\mathrm{m}\) and launches a \(0.5\,\mathrm{kg}\) ball horizontally across a frictionless table. Find the launch speed.
Given: \(k = 200\,\mathrm{N/m}\), \(x = 0.10\,\mathrm{m}\), \(m = 0.5\,\mathrm{kg}\), frictionless.
Find: the launch speed \(v\).
Setup: All the stored elastic energy converts to kinetic energy: \(\tfrac12 kx^2 = \tfrac12 mv^2\).
Solve:
\[ \tfrac12 kx^2 = \tfrac12 (200)(0.10)^2 = 1\,\mathrm{J} = \tfrac12 m v^2 \;\Rightarrow\; v = \sqrt{\frac{2(1)}{0.5}} = 2\,\mathrm{m/s}. \]
Answer: \(\boxed{v = 2\,\mathrm{m/s}}\)
Check: Compact form: \(v = x\sqrt{k/m} = 0.10\sqrt{200/0.5} = 0.10\sqrt{400} = 0.10(20) = 2\,\mathrm{m/s}\) ✓. The energy stored, \(1\,\mathrm{J}\), equals the kinetic energy delivered. ✓
27Power of a hoist
A motor lifts a \(50\,\mathrm{kg}\) load at a constant \(2\,\mathrm{m/s}\). Find (a) the power delivered and (b) the work done in \(8\,\mathrm{s}\).
Given: \(m = 50\,\mathrm{kg}\), \(v = 2\,\mathrm{m/s}\) (constant), \(t = 8\,\mathrm{s}\).
Find: power \(P\) and work \(W\) in \(8\,\mathrm{s}\).
Setup: At constant speed the motor's lifting force just balances weight, \(F = mg\), and \(P = Fv\).
Solve:
\[ P = Fv = mgv = (50)(10)(2) = 1000\,\mathrm{W} = 1\,\mathrm{kW}, \qquad W = Pt = 1000 \times 8 = 8000\,\mathrm{J}. \]
Answer: \(\boxed{P = 1\,\mathrm{kW},\quad W = 8\,\mathrm{kJ}}\)
Check: In \(8\,\mathrm{s}\) the load rises \(h = vt = 16\,\mathrm{m}\), so the gravitational PE gained is \(mgh = (50)(10)(16) = 8000\,\mathrm{J}\), matching the work done. ✓
28Reading equilibria off a potential
A particle moves along the \(x\)-axis with potential energy \(U(x) = x^3 - 3x\) (in joules, \(x\) in metres). Locate the equilibrium positions and classify each as stable or unstable.
Given: \(U(x) = x^3 - 3x\).
Find: equilibrium positions and their stability.
Setup: Equilibria occur where the force \(F = -dU/dx\) vanishes; the sign of \(U'' = d^2U/dx^2\) decides stability (minimum \(\Rightarrow\) stable, maximum \(\Rightarrow\) unstable).
Solve: \(F = -\dfrac{dU}{dx} = -(3x^2 - 3) = 3(1 - x^2)\), which is zero at \(x = \pm 1\).
\(\dfrac{d^2U}{dx^2} = 6x\): at \(x = +1\) it is \(+6 \gt 0\) (a minimum), at \(x = -1\) it is \(-6 \lt 0\) (a maximum).
Answer: \(\boxed{x = +1\,\mathrm{m}\ \text{stable},\quad x = -1\,\mathrm{m}\ \text{unstable}}\)
Check: Just past \(x = 1\) (say \(x = 1.1\)): \(F = 3(1 - 1.21) = -0.63\,\mathrm{N}\), pointing back toward \(x = 1\) — restoring, hence stable. Just past \(x = -1\) (say \(x = -0.9\)): \(F = 3(1 - 0.81) = +0.57\,\mathrm{N}\), pointing away from \(x = -1\) — hence unstable. ✓
JEE Main Style
29Spring launches a block up a rough incline
A spring (\(k = 500\,\mathrm{N/m}\)) at the foot of a \(37^\circ\) incline is compressed \(0.20\,\mathrm{m}\) and released, driving a \(1\,\mathrm{kg}\) block up the incline. The block–incline friction is \(\mu_k = 0.25\). How far up the incline (measured from the release point) does the block travel before stopping? (\(\sin37^\circ = 0.6,\ \cos37^\circ = 0.8\).)
Given: \(k = 500\,\mathrm{N/m}\), \(x = 0.20\,\mathrm{m}\), \(m = 1\,\mathrm{kg}\), \(\theta = 37^\circ\), \(\mu_k = 0.25\).
Find: the stopping distance \(L\) along the incline.
Setup: All the spring energy is spent climbing against gravity and grinding against friction over the same path length \(L\) (the incline angle, and so the normal force \(mg\cos\theta\), is constant throughout). Set spring PE equal to the sum of those two:
\[ \tfrac12 kx^2 = mgL\sin\theta + \mu_k mg\cos\theta\, L = mgL(\sin\theta + \mu_k\cos\theta). \]
Solve: \(\tfrac12 kx^2 = \tfrac12(500)(0.20)^2 = 10\,\mathrm{J}\), and \(\sin\theta + \mu_k\cos\theta = 0.6 + 0.25(0.8) = 0.8\), so
\[ 10 = (1)(10)\,L\,(0.8) = 8L \;\Rightarrow\; L = 1.25\,\mathrm{m}. \]
Answer: \(\boxed{L = 1.25\,\mathrm{m}}\)
Check: Energy budget: climb \(= mgL\sin\theta = (10)(1.25)(0.6) = 7.5\,\mathrm{J}\); friction \(= \mu_k mg\cos\theta\,L = (0.25)(10)(0.8)(1.25) = 2.5\,\mathrm{J}\); total \(= 10\,\mathrm{J}\), exactly the spring energy. ✓
30Vertical circle: the energy–Newton combo
A \(0.2\,\mathrm{kg}\) ball on a string of length \(L = 0.8\,\mathrm{m}\) is swung in a vertical circle. Find (a) the minimum speed at the top to keep the string taut, (b) the speed at the bottom for that case, and (c) the string tension at the bottom.
Given: \(m = 0.2\,\mathrm{kg}\), \(L = 0.8\,\mathrm{m}\).
Find: \(v_{\text{top,min}}\), \(v_{\text{bottom}}\), \(T_{\text{bottom}}\).
Setup: Newton fixes the top condition (at minimum, the string goes slack so gravity alone supplies the centripetal force); energy carries the speed from top to bottom; Newton again gives the bottom tension. This split — energy for speeds, \(\vec{F}=m\vec{a}\) for forces — is the whole trick.
Solve: At the top, \(mg = \dfrac{mv_{\text{top}}^2}{L} \Rightarrow v_{\text{top}}^2 = gL = (10)(0.8) = 8\,\mathrm{m^2/s^2}\), so \(v_{\text{top}} = 2\sqrt{2}\,\mathrm{m/s}\).
Energy from top to bottom (a drop of \(2L\)): \(v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4gL = 8 + 4(10)(0.8) = 40\), so \(v_{\text{bottom}} = 2\sqrt{10}\,\mathrm{m/s}\).
At the bottom, \(T - mg = \dfrac{mv_{\text{bottom}}^2}{L} \Rightarrow T = m\!\left(g + \dfrac{v_{\text{bottom}}^2}{L}\right) = 0.2\!\left(10 + \dfrac{40}{0.8}\right) = 0.2(60) = 12\,\mathrm{N}\).
Answer: \(\boxed{v_{\text{top}} = 2\sqrt{2}\,\mathrm{m/s},\quad v_{\text{bottom}} = 2\sqrt{10} \approx 6.3\,\mathrm{m/s},\quad T_{\text{bottom}} = 12\,\mathrm{N}}\)
Check: Standard result: for a vertical circle, \(T_{\text{bottom}} - T_{\text{top}} = 6mg\). Here \(T_{\text{top}} = 0\) (just taut), so \(T_{\text{bottom}} = 6mg = 6(0.2)(10) = 12\,\mathrm{N}\), matching. ✓
31Variable force by integration
A \(2\,\mathrm{kg}\) particle on a frictionless track starts from rest at \(x = 0\). A force \(F(x) = (6 - 2x)\,\mathrm{N}\) (with \(x\) in metres) acts along the track. Find (a) where the force reverses, (b) the speed at \(x = 2\,\mathrm{m}\), (c) the maximum speed and where it occurs, and (d) where the particle momentarily stops.
Given: \(m = 2\,\mathrm{kg}\), starts at rest at \(x = 0\), \(F(x) = 6 - 2x\).
Find: reversal point, \(v(2)\), \(v_{\max}\) and its location, and the turning point.
Setup: The work up to position \(x\) is the area under \(F(x)\): \(W(x) = \displaystyle\int_0^x (6 - 2x')\,dx' = 6x - x^2\). Then \(\tfrac12 mv^2 = W(x)\).
Solve: (a) \(F = 0\) at \(x = 3\,\mathrm{m}\) — the force pushes forward for \(x \lt 3\) and backward for \(x \gt 3\).
(b) \(W(2) = 12 - 4 = 8\,\mathrm{J} \Rightarrow v = \sqrt{8} = 2\sqrt{2}\,\mathrm{m/s}\).
(c) Kinetic energy is greatest where \(W(x)\) peaks, i.e. at \(x = 3\): \(W(3) = 18 - 9 = 9\,\mathrm{J} \Rightarrow v_{\max} = \sqrt{9} = 3\,\mathrm{m/s}\).
(d) The particle stops when \(W(x) = 0\) again: \(6x - x^2 = 0 \Rightarrow x = 6\,\mathrm{m}\).
Answer: \(\boxed{\text{reverses at } x=3\,\mathrm{m};\ v(2)=2\sqrt2\,\mathrm{m/s};\ v_{\max}=3\,\mathrm{m/s at } x=3;\ \text{stops at } x=6\,\mathrm{m}}\)
Check: \(W(x) = -(x-3)^2 + 9\) is a downward parabola peaking at \(x = 3\) and vanishing at \(x = 0\) and \(x = 6\) — symmetric about the force-reversal point, exactly as a turning-point picture demands. ✓
32A power-limited car
A \(1200\,\mathrm{kg}\) car has a maximum engine power of \(48\,\mathrm{kW}\) and meets a constant resistance of \(1200\,\mathrm{N}\). Find (a) its maximum speed on level road and (b) its acceleration at the instant its speed is \(20\,\mathrm{m/s}\).
Given: \(m = 1200\,\mathrm{kg}\), \(P = 48\,\mathrm{kW}\), resistance \(R = 1200\,\mathrm{N}\).
Find: \(v_{\max}\) and \(a\) at \(v = 20\,\mathrm{m/s}\).
Setup: The driving force is \(F = P/v\) (constant power), so it falls as the car speeds up. Top speed is reached when the driving force has dropped to the resistance; below that, the surplus accelerates the car.
Solve: (a) At \(v_{\max}\), \(F = R\): \(\dfrac{P}{v_{\max}} = R \Rightarrow v_{\max} = \dfrac{P}{R} = \dfrac{48000}{1200} = 40\,\mathrm{m/s}\).
(b) At \(v = 20\,\mathrm{m/s}\), \(F = P/v = 48000/20 = 2400\,\mathrm{N}\), so \(a = \dfrac{F - R}{m} = \dfrac{2400 - 1200}{1200} = 1\,\mathrm{m/s^2}\).
Answer: \(\boxed{v_{\max} = 40\,\mathrm{m/s},\quad a(20) = 1\,\mathrm{m/s^2}}\)
Check: At \(v = v_{\max} = 40\), \(F = 48000/40 = 1200\,\mathrm{N} = R\), giving \(a = 0\) — consistent with \(40\,\mathrm{m/s}\) being the top speed. ✓
33Pulley system by the energy method
Block \(A\) (\(2\,\mathrm{kg}\)) sits on a rough table (\(\mu_k = 0.5\)) and is joined over a light frictionless pulley to a hanging block \(B\) (\(3\,\mathrm{kg}\)). Released from rest, find the common speed after \(B\) has descended \(1\,\mathrm{m}\).
Given: \(m_A = 2\,\mathrm{kg}\) (table, \(\mu_k = 0.5\)), \(m_B = 3\,\mathrm{kg}\) (hanging), \(h = 1\,\mathrm{m}\), from rest.
Find: the speed \(v\) of the blocks.
Setup: Treat the two blocks (which share one speed) as a system. The PE that \(B\) loses is split between the system's kinetic energy and the heat made by friction on \(A\):
\[ m_B g h - \mu_k m_A g h = \tfrac12 (m_A + m_B) v^2. \]
Solve: Lost PE \(= (3)(10)(1) = 30\,\mathrm{J}\); friction \(= (0.5)(2)(10)(1) = 10\,\mathrm{J}\). So
\[ 30 - 10 = \tfrac12 (5) v^2 = 2.5\,v^2 \;\Rightarrow\; v^2 = 8 \;\Rightarrow\; v = 2\sqrt2\,\mathrm{m/s}. \]
Answer: \(\boxed{v = 2\sqrt2 \approx 2.8\,\mathrm{m/s}}\)
Check: Alternative method: \(a = \dfrac{m_B g - \mu_k m_A g}{m_A + m_B} = \dfrac{30 - 10}{5} = 4\,\mathrm{m/s^2}\), so \(v^2 = 2ah = 2(4)(1) = 8\), giving the same \(v = 2\sqrt2\,\mathrm{m/s}\). ✓
34Down a ramp, into a spring
A \(2\,\mathrm{kg}\) block is released from rest and descends a frictionless ramp through a height of \(1\,\mathrm{m}\), then slides along a frictionless floor into a spring of stiffness \(k = 1000\,\mathrm{N/m}\). Find (a) the maximum compression and (b) the block's speed when the spring is compressed half that amount.
Given: \(m = 2\,\mathrm{kg}\), \(h = 1\,\mathrm{m}\), frictionless, \(k = 1000\,\mathrm{N/m}\).
Find: maximum compression \(x\), and the speed at compression \(x/2\).
Setup: With no friction, the gravitational PE released sets the energy budget. At maximum compression the block is momentarily at rest, so all of it sits in the spring: \(mgh = \tfrac12 k x^2\). At a partial compression \(x'\), the budget splits between KE and spring PE.
Solve: (a) \(mgh = (2)(10)(1) = 20\,\mathrm{J} = \tfrac12 (1000) x^2 \Rightarrow x^2 = 0.04 \Rightarrow x = 0.2\,\mathrm{m}\).
(b) At \(x' = 0.1\,\mathrm{m}\): \(\tfrac12 mv^2 = mgh - \tfrac12 k x'^2 = 20 - \tfrac12(1000)(0.1)^2 = 20 - 5 = 15\,\mathrm{J}\), so \(v = \sqrt{2(15)/2} = \sqrt{15}\,\mathrm{m/s}\).
Answer: \(\boxed{x = 0.2\,\mathrm{m},\quad v(x/2) = \sqrt{15} \approx 3.9\,\mathrm{m/s}}\)
Check: Spring PE scales as \(x'^2\), so at half compression the spring holds \(\tfrac14\) of its maximum \(20\,\mathrm{J}\), i.e. \(5\,\mathrm{J}\), leaving \(15\,\mathrm{J}\) as KE — exactly as found. At \(x' = 0\) the speed would be \(\sqrt{2gh} = \sqrt{20} \approx 4.5\,\mathrm{m/s}\) (the maximum), and at \(x' = x\) it is zero. ✓
35Spring-launched projectile
A spring launcher (\(k = 800\,\mathrm{N/m}\), compressed \(0.10\,\mathrm{m}\)) fires a \(0.5\,\mathrm{kg}\) ball horizontally off the edge of a table \(1.25\,\mathrm{m}\) high. Find (a) the launch speed and (b) the horizontal distance from the table's foot where the ball lands.
Given: \(k = 800\,\mathrm{N/m}\), \(x = 0.10\,\mathrm{m}\), \(m = 0.5\,\mathrm{kg}\), \(H = 1.25\,\mathrm{m}\).
Find: launch speed \(v\) and range \(R\).
Setup: Energy gives the launch speed; then it is an ordinary horizontal-projectile problem — vertical fall sets the time, horizontal speed sets the range.
Solve: \(\tfrac12 kx^2 = \tfrac12(800)(0.10)^2 = 4\,\mathrm{J} = \tfrac12 mv^2 \Rightarrow v = \sqrt{2(4)/0.5} = \sqrt{16} = 4\,\mathrm{m/s}\).
Fall time: \(H = \tfrac12 g t^2 \Rightarrow t = \sqrt{2H/g} = \sqrt{2(1.25)/10} = \sqrt{0.25} = 0.5\,\mathrm{s}\).
Range: \(R = vt = 4 \times 0.5 = 2\,\mathrm{m}\).
Answer: \(\boxed{v = 4\,\mathrm{m/s},\quad R = 2\,\mathrm{m}}\)
Check: Dimensions: \(R = [\mathrm{m/s}][\mathrm{s}] = \mathrm{m}\) ✓. The launch is horizontal, so the vertical motion is unaffected by \(v\); doubling the spring energy would raise \(v\) by \(\sqrt2\) and so multiply \(R\) by \(\sqrt2\), not by \(2\). ✓
JEE Advanced Style
36Anatomy of a potential well
A particle of mass \(1\,\mathrm{kg}\) moves along the \(x\)-axis with potential energy \(U(x) = (x^2 - 1)^2\) joules (\(x\) in metres) and total mechanical energy \(E = 2\,\mathrm{J}\). Find (a) the equilibrium positions and their stability, (b) the turning points of the motion, (c) the maximum speed and where it occurs, and (d) the speed as it crosses \(x = 0\).
Given: \(m = 1\,\mathrm{kg}\), \(U(x) = (x^2 - 1)^2\), \(E = 2\,\mathrm{J}\).
Find: equilibria & stability; turning points; \(v_{\max}\) and its location; \(v(0)\).
Setup: Equilibria where \(U'(x) = 0\); stability from the sign of \(U''\). Turning points where \(U(x) = E\) (all KE spent). Speed anywhere from \(\tfrac12 mv^2 = E - U(x)\), maximal where \(U\) is least.
Solve: \(U'(x) = 4x(x^2 - 1)\), zero at \(x = 0\) and \(x = \pm1\). With \(U''(x) = 12x^2 - 4\): at \(x = \pm1\), \(U'' = 8 \gt 0\) (minima, \(U = 0\), stable); at \(x = 0\), \(U'' = -4 \lt 0\) (maximum, \(U = 1\,\mathrm{J}\), unstable) — a symmetric double well with a central barrier of height \(1\,\mathrm{J}\).
Turning points: \((x^2 - 1)^2 = 2 \Rightarrow x^2 - 1 = \pm\sqrt2 \Rightarrow x^2 = 1 + \sqrt2\) (the root \(1 - \sqrt2 \lt 0\) is rejected), so \(x = \pm\sqrt{1 + \sqrt2} \approx \pm1.55\,\mathrm{m}\).
Maximum speed at the well bottoms (\(x = \pm1\), \(U = 0\)): \(\tfrac12 v^2 = E - 0 = 2 \Rightarrow v_{\max} = 2\,\mathrm{m/s}\).
At \(x = 0\) (\(U = 1\,\mathrm{J}\)): \(\tfrac12 v^2 = 2 - 1 = 1 \Rightarrow v = \sqrt2\,\mathrm{m/s}\).
Answer: \(\boxed{x=\pm1\ \text{stable},\ x=0\ \text{unstable};\ \text{turning pts }\pm1.55\,\mathrm{m};\ v_{\max}=2\,\mathrm{m/s};\ v(0)=\sqrt2\,\mathrm{m/s}}\)
Check: Since \(E = 2\,\mathrm{J}\) exceeds the barrier height \(U(0) = 1\,\mathrm{J}\), the particle is not trapped in one well — it slows to \(\sqrt2\,\mathrm{m/s}\) at the barrier but crosses it and oscillates across the whole range \([-1.55,\,1.55]\,\mathrm{m}\), consistent with the single symmetric pair of turning points found. ✓
37Block on a movable wedge
A \(1\,\mathrm{kg}\) block is released from rest at the top of a frictionless wedge of mass \(3\,\mathrm{kg}\) and height \(0.6\,\mathrm{m}\). The wedge is free to slide on a frictionless floor. Find the speed of the block and of the wedge (both relative to the ground) when the block reaches the bottom.
Given: \(m = 1\,\mathrm{kg}\), \(M = 3\,\mathrm{kg}\), \(h = 0.6\,\mathrm{m}\), everything frictionless, from rest.
Find: the ground speeds \(V\) (block) and \(v_w\) (wedge) at the bottom.
Setup: No external horizontal force acts, and the system starts from rest, so horizontal momentum stays zero. At the bottom the incline meets the floor, so the block then moves horizontally — both bodies have horizontal velocities there. Momentum plus energy gives two equations.
Solve: Momentum: \(mV + Mv_w = 0 \Rightarrow v_w = -\dfrac{m}{M}V\) (wedge recoils opposite the block).
Energy: \(mgh = \tfrac12 mV^2 + \tfrac12 M v_w^2 = \tfrac12 mV^2\!\left(1 + \dfrac{m}{M}\right) = \tfrac12 mV^2\,\dfrac{M+m}{M}\), hence
\[ V^2 = \frac{2gh\,M}{M+m} = \frac{2(10)(0.6)(3)}{4} = 9 \;\Rightarrow\; V = 3\,\mathrm{m/s}, \qquad v_w = -\tfrac13(3) = -1\,\mathrm{m/s}. \]
Answer: \(\boxed{V = 3\,\mathrm{m/s (block)},\quad |v_w| = 1\,\mathrm{m/s (wedge, opposite)}}\)
Check: Momentum: \((1)(3) + (3)(-1) = 0\) ✓. Energy: \(\tfrac12(1)(9) + \tfrac12(3)(1) = 4.5 + 1.5 = 6\,\mathrm{J} = mgh = (1)(10)(0.6)\) ✓. (A fixed wedge would give the larger \(V = \sqrt{2gh} = \sqrt{12} \approx 3.46\,\mathrm{m/s}\); the recoiling wedge steals some energy, so \(V\) is smaller.) ✓
38Chain sliding off a table
A uniform chain of mass \(M\) and length \(L = 2\,\mathrm{m}\) lies on a frictionless table with a length \(\ell_0 = 0.5\,\mathrm{m}\) hanging over the edge. It is released. Find the speed of the chain at the instant the last link leaves the table.
Given: uniform chain, \(L = 2\,\mathrm{m}\), initial overhang \(\ell_0 = 0.5\,\mathrm{m}\), frictionless.
Find: the speed \(v\) when the chain just becomes fully vertical.
Setup: Energy is conserved; the work done by gravity equals the drop of the chain's centre of mass times \(Mg\). Measure depth below the table edge. (Mass cancels, as for a single particle.)
Solve: Initially the overhang (\(\ell_0\)) has its CM at depth \(\ell_0/2\) and the rest sits at the edge, so the whole chain's CM depth is \(y_i = \dfrac{\ell_0(\ell_0/2)}{L} = \dfrac{\ell_0^2}{2L}\). Finally the chain hangs vertically with CM depth \(y_f = L/2\). The CM drops by
\[ \Delta y = \frac{L}{2} - \frac{\ell_0^2}{2L} = \frac{L^2 - \ell_0^2}{2L}. \]
Energy: \(Mg\,\Delta y = \tfrac12 M v^2 \Rightarrow v^2 = g\,\dfrac{L^2 - \ell_0^2}{L} = (10)\dfrac{4 - 0.25}{2} = 18.75\), so
\[ v = \sqrt{18.75} = \frac{5\sqrt3}{2}\,\mathrm{m/s} \approx 4.33\,\mathrm{m/s}. \]
Answer: \(\boxed{v = \dfrac{5\sqrt3}{2} \approx 4.3\,\mathrm{m/s}}\)
Check: Limiting cases: if the chain already hung fully (\(\ell_0 \to L\)), then \(v^2 \to g(L^2 - L^2)/L = 0\) — nothing left to fall ✓; and the maximum possible speed (a vanishingly small initial overhang, \(\ell_0 \to 0\)) is \(\sqrt{gL}\), which the answer never exceeds ✓. Dimensions: \([gL]^{1/2} = \mathrm{m/s}\) ✓.
39Loop-the-loop with a rough patch
A block starts from rest on a smooth ramp of height \(h\). At the bottom it crosses a rough horizontal patch of length \(2\,\mathrm{m}\) (\(\mu_k = 0.5\)) and then enters a smooth vertical loop of radius \(R = 0.5\,\mathrm{m}\). Find the minimum \(h\) for which the block just completes the loop.
Given: smooth ramp, rough patch \(L = 2\,\mathrm{m}\) with \(\mu_k = 0.5\), smooth loop \(R = 0.5\,\mathrm{m}\), from rest.
Find: the minimum release height \(h\).
Setup: “Just completes the loop” means the speed at the top satisfies the minimum-contact condition \(v_{\text{top}}^2 = gR\) (gravity alone provides the centripetal force there). Then track the energy from release to the top of the loop, debiting the friction lost on the flat patch:
\[ mgh - \mu_k mg L = \tfrac12 m v_{\text{top}}^2 + mg(2R). \]
Solve: Divide by \(mg\) and use \(v_{\text{top}}^2 = gR\):
\[ h - \mu_k L = \tfrac12 R + 2R = \tfrac{5}{2}R \;\Rightarrow\; h = \mu_k L + \tfrac{5}{2}R = (0.5)(2) + (2.5)(0.5) = 1 + 1.25 = 2.25\,\mathrm{m}. \]
Answer: \(\boxed{h_{\min} = 2.25\,\mathrm{m}}\)
Check: With no rough patch (\(\mu_k = 0\)) the formula collapses to the textbook \(h = \tfrac52 R = 1.25\,\mathrm{m}\) ✓. Friction can only raise the required height, and it does (\(2.25 \gt 1.25\,\mathrm{m}\)). ✓
40Pump with efficiency
A pump draws water from \(10\,\mathrm{m}\) below ground and ejects it at the surface through a nozzle at \(4\,\mathrm{m/s}\), delivering \(30\,\mathrm{kg}\) of water each second. The pump is \(80\%\) efficient. Find the electrical power it must draw.
Given: lift \(h = 10\,\mathrm{m}\), exit speed \(v = 4\,\mathrm{m/s}\), mass rate \(\dot m = 30\,\mathrm{kg/s}\), efficiency \(\eta = 0.8\).
Find: input power \(P_{\text{in}}\).
Setup: The useful power is the rate at which the water gains energy — both potential (raised by \(h\)) and kinetic (leaves at speed \(v\)): \(P_{\text{useful}} = \dot m\!\left(gh + \tfrac12 v^2\right)\). Then \(P_{\text{in}} = P_{\text{useful}}/\eta\).
Solve:
\[ P_{\text{useful}} = 30\!\left[(10)(10) + \tfrac12(4)^2\right] = 30\,[100 + 8] = 3240\,\mathrm{W}, \qquad P_{\text{in}} = \frac{3240}{0.8} = 4050\,\mathrm{W}. \]
Answer: \(\boxed{P_{\text{in}} = 4050\,\mathrm{W} \approx 4.05\,\mathrm{kW}}\)
Check: The PE term dominates (\(3000\,\mathrm{W}\)) over the KE term (\(240\,\mathrm{W}\)), which is sensible for a \(10\,\mathrm{m}\) lift at a modest \(4\,\mathrm{m/s}\); and the \(20\%\) inefficiency adds \(810\,\mathrm{W}\) of waste, recovering \(P_{\text{useful}} = 0.8 \times 4050 = 3240\,\mathrm{W}\). ✓
41Block on a plank: where does the energy go?
A \(2\,\mathrm{kg}\) block sits on a \(4\,\mathrm{kg}\) plank that rests on a frictionless floor. A constant horizontal force \(F = 12\,\mathrm{N}\) is applied to the block; the block–plank friction is \(\mu_k = 0.2\). After the block has slid \(1\,\mathrm{m}\) relative to the plank, find (a) the heat generated, (b) the work done by \(F\), and verify the system energy balance.
Given: block \(m = 2\,\mathrm{kg}\), plank \(M = 4\,\mathrm{kg}\), frictionless floor, \(\mu_k = 0.2\), \(F = 12\,\mathrm{N}\), relative slip \(s_{\text{rel}} = 1\,\mathrm{m}\).
Find: heat \(Q\), work \(W_F\), and a check that \(W_F = \Delta K_{\text{sys}} + Q\).
Setup: First confirm they slide (not move together): if locked, \(a = F/(m+M) = 2\,\mathrm{m/s^2}\) would need \(Ma = 8\,\mathrm{N}\) on the plank, but friction can supply at most \(\mu_k mg = 4\,\mathrm{N}\) — so they slip. Friction acts backward on the block and forward on the plank.
Solve: \(f = \mu_k mg = (0.2)(2)(10) = 4\,\mathrm{N}\).
\[ a_{\text{block}} = \frac{F - f}{m} = \frac{12 - 4}{2} = 4\,\mathrm{m/s^2}, \qquad a_{\text{plank}} = \frac{f}{M} = \frac{4}{4} = 1\,\mathrm{m/s^2}. \]
Relative acceleration \(= 3\,\mathrm{m/s^2}\); reaching \(s_{\text{rel}} = 1\,\mathrm{m}\) takes \(t = \sqrt{2 s_{\text{rel}}/a_{\text{rel}}} = \sqrt{2/3}\,\mathrm{s}\), so \(t^2 = \tfrac23\,\mathrm{s^2}\). Then the block advances \(s_{\text{block}} = \tfrac12 a_{\text{block}} t^2 = \tfrac12(4)(\tfrac23) = \tfrac43\,\mathrm{m}\) (and the plank \(\tfrac13\,\mathrm{m}\), so their difference is the \(1\,\mathrm{m}\) slip).
(a) Heat is friction force times relative sliding: \(Q = f\,s_{\text{rel}} = 4 \times 1 = 4\,\mathrm{J}\).
(b) Work by \(F\) (which moves with the block): \(W_F = F\,s_{\text{block}} = 12 \times \tfrac43 = 16\,\mathrm{J}\).
Answer: \(\boxed{Q = 4\,\mathrm{J},\quad W_F = 16\,\mathrm{J}}\)
Check: Kinetic energies: \(v_{\text{block}} = a_{\text{block}}t\), \(v_{\text{plank}} = a_{\text{plank}}t\), giving \(K_{\text{block}} = \tfrac12(2)(16)(\tfrac23) = \tfrac{32}{3}\,\mathrm{J}\) and \(K_{\text{plank}} = \tfrac12(4)(1)(\tfrac23) = \tfrac43\,\mathrm{J}\), so \(\Delta K_{\text{sys}} = 12\,\mathrm{J}\). Then \(\Delta K_{\text{sys}} + Q = 12 + 4 = 16\,\mathrm{J} = W_F\) — the energy input is fully accounted for, the missing \(4\,\mathrm{J}\) being heat from internal sliding. ✓
42Why the work–energy theorem holds for any force
Prove that for a particle moving along the \(x\)-axis under a net force \(F(x)\) — constant or variable — the work done equals the change in kinetic energy.
Setup: Start from Newton's second law and rewrite the acceleration as a derivative with respect to position, so the whole equation can be integrated over the displacement.
Solve: By the chain rule,
\[ a = \frac{dv}{dt} = \frac{dv}{dx}\,\frac{dx}{dt} = v\,\frac{dv}{dx}. \]
Newton's second law \(F = ma\) becomes \(F = m v\,\dfrac{dv}{dx}\), i.e. \(F\,dx = m v\,dv\). Integrating from the initial state to the final state,
\[ \int_{x_i}^{x_f} F\,dx = \int_{v_i}^{v_f} m v\,dv = \tfrac12 m v_f^2 - \tfrac12 m v_i^2. \]
The left side is the work \(W\) done by the net force; the right side is \(\Delta K\). Hence \(W = \Delta K\).
Answer: \(\boxed{\displaystyle \int_{x_i}^{x_f} F\,dx = \tfrac12 m v_f^2 - \tfrac12 m v_i^2}\)
Check: Nowhere was \(F\) assumed constant, so the result covers variable forces. As a special case, a constant \(F\) acting from rest over a distance \(d\) gives \(Fd = \tfrac12 mv^2\), i.e. \(v = \sqrt{2Fd/m}\) — exactly the kinematic \(v^2 = 2(F/m)d\) ✓. In three dimensions the identical steps with \(\vec F\cdot d\vec r = m\,\vec v\cdot d\vec v\) reproduce the theorem with \(v^2 = \vec v\cdot\vec v\).