"Work, Energy and Power" — Practice

Question 1

A block slides on a plank that is itself free to move. Explain why the heat generated is not the friction force times the block's displacement.

Accepted Answer

Friction does $-fs_1$ on the block and $+fs_2$ on the plank, where $s_1$ and $s_2$ are their **separate** displacements.

Total work by the pair $=-fs_1+fs_2=-f\left(s_1-s_2\right)=-f\,s_{rel}$

$$Q=f\,s_{rel}$$

Using $fs_1$ instead computes the work done **on the block**, which is a different quantity — part of the block's lost energy went into speeding the plank up, not into heat.

*When the two agree:* if the lower surface is fixed, $s_2=0$ and $s_{rel}=s_1$. Every elementary friction problem is of that kind, which is why the distinction is so easily missed.

Question 2

A particle moves under $U(x)=x^{3}-6x^{2}+9x$ (SI units). Find the equilibrium positions and classify them.

Accepted Answer

$\dfrac{dU}{dx}=3x^{2}-12x+9=3(x-1)(x-3)$

Equilibria at $x=\mathbf{1}$ and $x=\mathbf{3}$.

$\dfrac{d^{2}U}{dx^{2}}=6x-12$

| $x$ | $U''$ | Nature |
|---|---|---|
| $1$ | $-6$ | maximum, **unstable** |
| $3$ | $+6$ | minimum, **stable** |

*Check with the force directly:* at $x=3.1$, $F=-3(2.1)(0.1)=-0.63$ N, pushing back toward $x=3$. Restoring, so stable, as the second derivative said.

Question 3

Two equal masses collide head-on and stick together, one having been at rest. What fraction of the kinetic energy is lost?

Accepted Answer

$v_{cm}=\dfrac{mu}{2m}=\dfrac{u}{2}$

$K_i=	frac12mu^{2}, \qquad K_f=	frac12(2m)\left(\dfrac{u}{2}ight)^{2}=	frac14mu^{2}$

$$	ext{Fraction lost}=\mathbf{50\%}$$

*Via the reduced mass, in one line:* the available energy is $	frac12\mu u^{2}$ with $\mu=\dfrac{m\cdot m}{2m}=\dfrac{m}{2}$, so the loss is $	frac12\left(\dfrac{m}{2}ight)u^{2}=	frac14mu^{2}$, which is half of $K_i$.

*Why not 100%?* Momentum conservation forces the pair to keep moving at $u/2$, and that translational kinetic energy is untouchable.

Question 4

A $1\ 	ext{kg}$ block slides at $8\ 	ext{m/s}$ onto a $3\ 	ext{kg}$ plank on frictionless ice, with $\mu=0.4$ between them and $g=10$. Find the heat generated and the distance the block slides along the plank.

Accepted Answer

$v_{cm}=\dfrac{1(8)}{4}=2\ 	ext{m/s}$

$K_i=	frac12(1)(64)=32\ 	ext{J}, \qquad K_f=	frac12(4)(4)=8\ 	ext{J}$

$$Q=\mathbf{24\ 	ext{J}}$$

*Check with the reduced mass:* $\mu_{red}=\dfrac{1(3)}{4}=0.75$, so $Q=	frac12(0.75)(64)=24$ J.

**Now the sliding distance**, which is where $\mu$ finally matters:

$f=\mu m g=0.4(1)(10)=4\ 	ext{N}$

$s_{rel}=\dfrac{Q}{f}=\dfrac{24}{4}=\mathbf{6\ 	ext{m}}$

*The plank must therefore be at least 6 m long*, or the block slides off the end and the calculation stops early — a favourite twist.

Question 5

A particle moves under $U(x,y)=2x^{2}y+3y^{2}$ (SI units). Find the force at the point $(2,1)$.

Accepted Answer

$F_x=-\dfrac{\partial U}{\partial x}=-4xy$, so at $(2,1)$: $F_x=-8\ 	ext{N}$

$F_y=-\dfrac{\partial U}{\partial y}=-\left(2x^{2}+6yight)$, so at $(2,1)$: $F_y=-(8+6)=-14\ 	ext{N}$

$$\vec{F}=(-8,\ -14)\ 	ext{N}, \qquad |\vec{F}|=\sqrt{64+196}=\mathbf{16.1\ 	ext{N}}$$

*Direction check:* both components are negative, so the force points toward the third quadrant — down the potential hill, which is what $-
abla U$ always does.

*Note:* $(2,1)$ is not an equilibrium, since neither component vanishes. Equilibrium would require $xy=0$ **and** $2x^{2}+6y=0$ simultaneously.

Question 6

A $0.2\ 	ext{kg}$ particle moves under $U(x)=\dfrac{8}{x^{2}}-\dfrac{8}{x}$ in SI units. Find the stable equilibrium and the period of small oscillations about it.

Accepted Answer

$\dfrac{dU}{dx}=-\dfrac{16}{x^{3}}+\dfrac{8}{x^{2}}=0 \Rightarrow 8x=16 \Rightarrow x_0=\mathbf{2\ 	ext{m}}$

$\dfrac{d^{2}U}{dx^{2}}=\dfrac{48}{x^{4}}-\dfrac{16}{x^{3}}$

At $x_0=2$: $\dfrac{48}{16}-\dfrac{16}{8}=3-2=1\ 	ext{N/m}$

Positive, so **stable**, and the effective stiffness is $1$ N/m.

$\omega=\sqrt{\dfrac{1}{0.2}}=2.24\ 	ext{rad/s} \Rightarrow T=\dfrac{2\pi}{\omega}=\mathbf{2.81\ 	ext{s}}$

*Recognise the potential.* An inverse-square repulsion plus an inverse attraction is the standard model of a chemical bond, and this calculation is exactly how a molecule's vibrational frequency is estimated from its potential curve.

Question 7

Blocks of $1\ 	ext{kg}$ and $4\ 	ext{kg}$ on a frictionless floor are joined by a spring of stiffness $500\ 	ext{N/m}$ at natural length. The $1\ 	ext{kg}$ block is given $10\ 	ext{m/s}$ toward the other. Find the maximum compression.

Accepted Answer

Maximum compression occurs when the separation stops changing, that is when both blocks share a velocity:

$v=\dfrac{1(10)}{5}=2\ 	ext{m/s}$

$	frac12(1)(100)=	frac12(5)(4)+	frac12(500)x^{2}$

$50=10+250x^{2} \Rightarrow x^{2}=0.16 \Rightarrow x=\mathbf{0.4\ 	ext{m}}$

*One-line route.* The energy stored is exactly the internal kinetic energy, $	frac12\mu v_{rel}^{2}$ with $\mu=\dfrac{1(4)}{5}=0.8$:

$	frac12(0.8)(100)=40\ 	ext{J}=	frac12(500)x^{2} \Rightarrow x=0.4\ 	ext{m}$

This route also shows **why** the answer is frame-independent: the translational energy $	frac12(5)(4)=10$ J never had any way of reaching the spring.

Question 8

A ball strikes a smooth horizontal floor at $60°$ to the horizontal with speed $u$. The coefficient of restitution is $0.4$. Find the rebound speed and angle.

Accepted Answer

The line of impact is **vertical**, so only the vertical component is reduced.

| Component | Before | After |
|---|---|---|
| Horizontal | $u\cos 60°=0.5u$ | unchanged, $0.5u$ |
| Vertical | $u\sin 60°=0.866u$ | $0.4(0.866u)=0.346u$ |

$v=\sqrt{(0.5u)^{2}+(0.346u)^{2}}=\mathbf{0.608u}$

$	an	heta'=\dfrac{0.346}{0.5}=0.693 \Rightarrow 	heta'=\mathbf{34.7°}$

*The ball leaves much flatter than it arrived* — $35°$ against $60°$ — because only the vertical component was cut. Repeated bounces flatten the angle further each time, which is why a bouncing ball eventually skids rather than bounces.

Question 9

A wedge of angle $	heta$ accelerates horizontally at $a$, in the direction that makes its incline descend forward. A block slides down the frictionless incline. Working **in the wedge's frame**, find its speed relative to the wedge after sliding a distance $L$ along the incline.

Accepted Answer

In the wedge's frame the block starts at rest, and two forces have components along the incline. The normal force does no work, being perpendicular to the motion.

**Gravity**, component down the slope: $mg\sin	heta$

**Pseudo force** $ma$ backward, component along the slope: $-ma\cos	heta$

Work-energy theorem over a slide of length $L$:

$\left(mg\sin	heta-ma\cos	hetaight)L=	frac12mv_{rel}^{2}$

$$v_{rel}=\sqrt{2L\left(g\sin	heta-a\cos	hetaight)}$$

*Two checks.* Setting $a=0$ recovers $\sqrt{2gL\sin	heta}$, the fixed-incline result.

Setting $a=g	an	heta$ makes the bracket vanish, so $v_{rel}=0$ — the block never slides at all. That is exactly the non-sliding condition derived in Laws of Motion, recovered here from energy instead of forces.

Question 10

A $2\ 	ext{kg}$ ball moving at $9\ 	ext{m/s}$ collides elastically and head-on with a stationary $4\ 	ext{kg}$ ball. Solve for both final velocities using the centre-of-mass frame.

Accepted Answer

$v_{cm}=\dfrac{2(9)}{6}=3\ 	ext{m/s}$

**Transform into the CM frame** by subtracting $v_{cm}$:

$u_1'=9-3=6\ 	ext{m/s}, \qquad u_2'=0-3=-3\ 	ext{m/s}$

*Check:* $2(6)+4(-3)=0$, as the CM frame requires.

**Elastic collision in this frame simply reverses both:**

$v_1'=-6, \qquad v_2'=+3$

**Transform back** by adding $v_{cm}$:

$v_1=-6+3=\mathbf{-3\ 	ext{m/s}}, \qquad v_2=3+3=\mathbf{6\ 	ext{m/s}}$

*Check against the standard formula:* $v_1=\dfrac{2-4}{6}(9)=-3$ and $v_2=\dfrac{2(2)}{6}(9)=6$. Identical.

The CM route required no formula at all — only the fact that an elastic collision in that frame preserves both speeds.

Question 11

A particle of mass $m$ moving at $u$ strikes a stationary particle of mass $M$. Find the maximum possible fraction of kinetic energy that can be lost, state when it occurs, and explain why 100% is impossible.

Accepted Answer

By König's theorem the energy splits into an untouchable translational part and an available internal part:

$K_{internal}=	frac12\mu u^{2}, \qquad \mu=\dfrac{mM}{m+M}$

The maximum loss is all of it, which happens when the collision is **perfectly inelastic** ($e=0$) and the two move off together.

$$	ext{Fraction lost}=\frac{	frac12\mu u^{2}}{	frac12mu^{2}}=\frac{\mu}{m}=\mathbf{\frac{M}{m+M}}$$

**Why never 100%.** Momentum conservation forces the system to keep moving at $v_{cm}=\dfrac{mu}{m+M}$, and the associated $	frac12(m+M)v_{cm}^{2}$ cannot be destroyed by any internal process.

| Case | Fraction lost |
|---|---|
| $M=m$ | $50\%$ |
| $M\gg m$ | approaches $100\%$ |
| $M\ll m$ | approaches $0\%$ |

*The physical reading:* a light body hitting a heavy one loses almost everything, which is why neutrons are moderated by hydrogen and not by lead.

Question 12

A $1\ 	ext{kg}$ block slides from rest down a frictionless quarter-circle track of radius $1.25\ 	ext{m}$ and lands on a $3\ 	ext{kg}$ plank resting on frictionless ice. The coefficient of friction between block and plank is $0.2$. Take $g=10$. Find the final common velocity, the heat generated, and the relative sliding distance.

Accepted Answer

**Stage 1 — down the track.** Frictionless, so energy is conserved:

$v=\sqrt{2gh}=\sqrt{2(10)(1.25)}=5\ 	ext{m/s}$

**Stage 2 — onto the plank.** The ice is frictionless, so horizontal momentum is conserved:

$v_{cm}=\dfrac{1(5)}{4}=\mathbf{1.25\ 	ext{m/s}}$

**Stage 3 — the heat.**

$K_i=	frac12(1)(25)=12.5\ 	ext{J}, \qquad K_f=	frac12(4)(1.5625)=3.125\ 	ext{J}$

$$Q=\mathbf{9.375\ 	ext{J}}$$

*Reduced-mass check:* $\mu_{red}=\dfrac{1(3)}{4}=0.75$, giving $	frac12(0.75)(25)=9.375$ J.

**Sliding distance:** $f=0.2(1)(10)=2\ 	ext{N}$, so

$s_{rel}=\dfrac{9.375}{2}=\mathbf{4.69\ 	ext{m}}$

*Note which stage needed which law.* Energy on the frictionless track, momentum during the transfer, and only then energy again for the heat. Using energy conservation across stage 2 would have been wrong, because friction is dissipative there.

Practice — "Work, Energy and Power"

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