Work, Energy and Power

Assume the box moves slowly (at constant velocity), so the pulling force just balances the sliding friction.

Normal reaction $N=mg=15\times9.8=147$ N.

Friction $f=\mu N=0.3\times147=44.1$ N, so the pulling force is $F=44.1$ N.

Work done for a distance $d=1$ m (unit distance):

$W=Fd=44.1\times1=44.1\ \text{J}$

The work done by the force is about $44\ \text{J}$ for each metre moved (45 J if $g=10\ \text{m s}^{-2}$ is used).

The lighter body has the greater kinetic energy.

Kinetic energy can be written in terms of momentum $p$ and mass $m$:

$K=\frac{1}{2}mv^2=\frac{p^2}{2m}$

For the same $p$, $K\propto \dfrac{1}{m}$, so the body with the smaller mass has the larger kinetic energy.

(c) 5:4

Kinetic energy in terms of momentum is $K=\dfrac{p^2}{2m}$. For equal $p$, $K\propto \dfrac1m$, so

$\frac{K_1}{K_2}=\frac{m_2}{m_1}=\frac{5}{4}$

By the work–energy theorem, the work done by the opposing force equals the change in kinetic energy:

$W=-Fs=-100\times2=-200\ \text{J}$ $K_B=K_A+W=800-200=600\ \text{J}$

The kinetic energy at B is $600\ \text{J}$.

(c) $v\propto 1/\sqrt{m}$

By the work-energy theorem, $Fd=\tfrac12mv^2$, so $v=\sqrt{2Fd/m}$. With $F$ and $d$ fixed, $v\propto 1/\sqrt m$.

(b) negative

Kinetic friction opposes the relative motion, so it generally acts opposite to the displacement of the body. Then $W=fs\cos180^\circ=-fs$ is negative.

Take $g=9.81\ \text{m s}^{-2}$. Masses: bullet $m=2\ \text{g}=0.002\ \text{kg}$, block $M=0.600\ \text{kg}$, together $0.602\ \text{kg}$; rise $h=0.086\ \text{m}$.

(a) $\Delta E_p=(m+M)gh=0.602\times9.81\times0.086\approx0.51\ \text{J}$

(b) Just after impact the combined kinetic energy changes entirely into this potential energy: $\tfrac12(m+M)V^2=0.51\ \text{J}\;\Rightarrow\;V=\sqrt{2gh}=\sqrt{2\times9.81\times0.086}\approx1.30\ \text{m s}^{-1}$ Momentum is conserved in the impact: $mu=(m+M)V\;\Rightarrow\;u=\frac{0.602\times1.30}{0.002}\approx3.9\times10^{2}\ \text{m s}^{-1}$ (about $391\ \text{m s}^{-1}$).

(c) $E_k=\tfrac12mu^2=\tfrac12\times0.002\times(391)^2\approx1.5\times10^{2}\ \text{J}$

(d) The kinetic energy before impact (about 153 J) is far greater than the energy of the block and bullet afterwards (about 0.51 J). Almost all of it has been converted to heat, sound and deformation, so kinetic energy is not conserved. Momentum is conserved, but the bullet embeds in the block and they move together, so the collision is perfectly inelastic.

(a) The bird has kinetic energy because of its motion, and gravitational potential energy because of its height above the ground.

(b) Let a body of mass $m$ with initial velocity $u$ be acted on by a constant force $F$, giving it acceleration $a$ over a displacement $s$ so that its velocity becomes $v$. Then $v^2=u^2+2as$ Multiply by $\tfrac12m$: $\tfrac12mv^2=\tfrac12mu^2+(ma)s$ Since $F=ma$ and $W=Fs$, $\tfrac12mv^2=\tfrac12mu^2+W$ that is, final kinetic energy = initial kinetic energy + work done on the body.

(c) The winning team pulls the rope forward and does positive work on the rope, since the rope moves in the direction of its pull. The team giving way exerts a force opposite to the displacement, so it does negative work (equal in size to the work done by the winning team if the speed is constant).

(a) Power is the rate of doing work:

$P=\frac{W}{t}\quad\text{i.e.}\quad W=Pt$

(b) With $P=200\ \text{W}$ and $t=1800\ \text{s}$:

$W=Pt=200\times1800=3.6\times10^{5}\ \text{J}$

(b) 60 MW

Input power from falling water: $P_{\text{in}}=\frac{mgh}{t}=\frac{780000\times 9.8\times 550}{60}=7.0\times 10^7\ \text{W}\approx 70\ \text{MW}$

Output power at 85% efficiency: $P_{\text{out}}=0.85\times 70\ \text{MW}\approx 60\ \text{MW}$