"Work, Energy and Power" — Practice

Question 1

A force of $20\ 	ext{N}$ acts at $60°$ to a displacement of $5\ 	ext{m}$. Find the work done.

Accepted Answer

$W=Fd\cos	heta=20	imes5	imes\cos 60°=20	imes5	imes0.5=\mathbf{50\ 	ext{J}}$

Half of what the same force would do if aligned with the motion — the perpendicular component contributes nothing.

Question 2

Name three situations in which a force does zero work despite acting on a moving body.

Accepted Answer

1. **Centripetal force** in uniform circular motion.
2. **Normal force** on a body sliding along a surface.
3. **Tension** in the string of a conical pendulum or any circular motion.

In every case the force is perpendicular to the displacement, so $\cos 90°=0$.

A fourth, slightly different case: any force on a **stationary** body does no work, however large.

Question 3

Two equal masses collide elastically head-on, one moving and one at rest. Describe the outcome.

Accepted Answer

With $m_1=m_2$ the coefficient $\dfrac{m_1-m_2}{m_1+m_2}$ vanishes and $\dfrac{2m_2}{m_1+m_2}=1$.

**They exchange velocities exactly** — the moving one stops dead, the stationary one departs with the original speed.

This is why a cue ball stops on a square full-ball hit, and it is the single most-set special case in the topic.

Question 4

A force $F=3x^{2}+2x$ newtons acts from $x=0$ to $x=2\ 	ext{m}$. Find the work done.

Accepted Answer

$W=\displaystyle\int_0^2 (3x^{2}+2x)\,dx=\big[x^{3}+x^{2}\big]_0^2=8+4=\mathbf{12\ 	ext{J}}$

The tempting shortcut — take the midpoint force $F(1)=5\ 	ext{N}$ and multiply by 2 m — gives 10 J and is **wrong**. Averaging the force only works when $F$ is linear in $x$.

Question 5

A 2 kg block slides $10\ 	ext{m}$ down a $30°$ incline from rest, $\mu_k=0.2$, $g=10$. Find its speed at the bottom.

Accepted Answer

Height dropped: $h=10\sin 30°=5\ 	ext{m}$, so gravity does $mgh=2(10)(5)=100\ 	ext{J}$.

Friction: $f_k=\mu_k mg\cos 30°=0.2(2)(10)(0.866)=3.46\ 	ext{N}$, doing $-3.46(10)=-34.6\ 	ext{J}$ over the path.

$W_{net}=100-34.6=65.4\ 	ext{J}=	frac12(2)v^{2}$

$v\approx\mathbf{8.1\ 	ext{m/s}}$

*Check:* frictionless would give $\sqrt{2(10)(5)}=10\ 	ext{m/s}$. Friction reduced it, as it must.

Question 6

A 2 kg ball at $6\ 	ext{m/s}$ strikes a stationary 4 kg ball elastically head-on. Find both final velocities and verify both conservation laws.

Accepted Answer

$v_1=\dfrac{2-4}{6}(6)=\mathbf{-2\ 	ext{m/s}}$ — it rebounds, as a lighter body must.

$v_2=\dfrac{2(4)}{6}(6)=\mathbf{4\ 	ext{m/s}}$

*Momentum:* before $2(6)=12$; after $2(-2)+4(4)=12$.

*Kinetic energy:* before $	frac12(2)(36)=36\ 	ext{J}$; after $	frac12(2)(4)+	frac12(4)(16)=4+32=36\ 	ext{J}$.

Both hold, confirming the collision is elastic.

Question 7

A pump raises $600\ 	ext{kg}$ of water per minute through $20\ 	ext{m}$ and delivers it at $5\ 	ext{m/s}$, $g=10$. Find the power required.

Accepted Answer

Flow rate $=\dfrac{600}{60}=10\ 	ext{kg/s}$.

**Potential:** $\dot{m}gh=10(10)(20)=2000\ 	ext{W}$

**Kinetic:** $	frac12\dot{m}v^{2}=	frac12(10)(25)=125\ 	ext{W}$

Total $\approx\mathbf{2125\ 	ext{W}}$

Forgetting the kinetic term is the standard error — the water leaves moving, not at rest.

Question 8

A ball on a string of length $L$ just completes a vertical circle. Derive its speed at the bottom.

Accepted Answer

At the top, "just completes" means the tension has fallen to zero, so **gravity alone** supplies the centripetal force:

$mg=\dfrac{mv_{top}^{2}}{L}\Rightarrow v_{top}^{2}=gL$

Energy conservation from top to bottom, a drop of $2L$:

$	frac12 v_{bot}^{2}=	frac12 v_{top}^{2}+g(2L)$

$v_{bot}^{2}=gL+4gL=5gL\Rightarrow v_{bot}=\mathbf{\sqrt{5gL}}$

The string does no work throughout — it is always perpendicular to the motion.

Question 9

A ball dropped from height $h$ bounces with coefficient of restitution $e$. Find the height after the $n$-th bounce and the total distance travelled.

Accepted Answer

Speed multiplies by $e$, and $h\propto v^{2}$, so each bounce multiplies the **height by $e^{2}$**:

$h_n=e^{2n}h$

Total distance: the first drop $h$, then every later height is covered twice (up and down):

$d=h+2h\left(e^{2}+e^{4}+\cdotsight)=h+2h\dfrac{e^{2}}{1-e^{2}}$

$d=h\dfrac{1+e^{2}}{1-e^{2}}$

*Check the limits:* $e=0$ gives $d=h$ (no bounce), and $e	o1$ makes $d$ diverge — a perfectly elastic ball never stops.

Question 10

Given a potential-energy curve $U(x)$, explain how to identify the stable equilibria and the limits of the motion for a given total energy.

Accepted Answer

**Equilibria:** points of zero slope, since $F=-dU/dx$.

**Stability:** a **minimum** is stable — a small displacement produces a restoring force back toward it. A **maximum** is unstable. A flat region is neutral.

**Limits of motion:** draw the total energy $E$ as a horizontal line. Since $K=E-U$ and $K\ge0$, the body can only be where the curve lies **below** the line.

**Turning points:** where the line meets the curve. There $K=0$, the body is momentarily at rest and reverses.

Trapped between two turning points is **bound** motion; only one turning point means the body escapes — which is exactly the escape-velocity condition in gravitation.

Question 11

A 2 kg block moving at $5\ 	ext{m/s}$ strikes a stationary 3 kg block and the two stick together. Find the common velocity and the percentage of kinetic energy lost.

Accepted Answer

Momentum is conserved, kinetic energy is not:

$v=\dfrac{2(5)+3(0)}{2+3}=\mathbf{2\ 	ext{m/s}}$

$K_i=	frac12(2)(25)=25\ 	ext{J}$, $K_f=	frac12(5)(4)=10\ 	ext{J}$

Lost $=15\ 	ext{J}$, i.e. $\mathbf{60\%}$.

*Check against the standard result:* with the target at rest the fraction lost is $\dfrac{m_2}{m_1+m_2}=\dfrac{3}{5}=60\%$. Agrees.

Question 12

A ball moving at $10\ 	ext{m/s}$ strikes an identical stationary ball elastically and moves off at $30°$ to its original direction. Find both final speeds and the direction of the second ball.

Accepted Answer

Momentum: $\vec{u}=\vec{v}_1+\vec{v}_2$. Energy: $u^{2}=v_1^{2}+v_2^{2}$.

Squaring the first gives $u^{2}=v_1^{2}+v_2^{2}+2\vec{v}_1\cdot\vec{v}_2$, so $\vec{v}_1\cdot\vec{v}_2=0$ — **they separate at $90°$**. The second ball therefore goes off at $\mathbf{60°}$ on the other side.

Along the original direction: $10=v_1\cos30°+v_2\cos60°$

Perpendicular to it: $0=v_1\sin30°-v_2\sin60°\Rightarrow v_1=1.732\,v_2$

Substituting: $10=1.5v_2+0.5v_2=2v_2$

$v_2=\mathbf{5\ 	ext{m/s}},\qquad v_1=\mathbf{8.66\ 	ext{m/s}}$

*Check energy:* $100$ before; $75+25=100$ after. Elastic confirmed.

Practice — "Work, Energy and Power"

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