Chapter 7: Work, Energy, and Simple Machines (Physics)

(i) False. Work is done only when the object is displaced; with zero displacement, $W=F\times 0=0$.

(ii) True. The lifting force and the displacement are both upward.

(iii) True. Work is the transfer of energy, so both are measured in joules.

(iv) False. It is not moving, so it has no kinetic energy; it has elastic potential energy because it is stretched.

(v) True. For example, electrical energy changes into light and heat in a bulb.

(i) F (ii) T (iii) T (iv) F (v) T

(i) Force × displacement (ii) 1 newton (iii) $\frac{1}{2}m{v}^2$ (iv) $mgh$ (v) rate

(i) force; displacement (ii) 1 (iii) ½mv² (iv) mgh (v) rate

Correct: (iii) and (iv).

At the highest point the ball is momentarily at rest, so its KE is zero, and it is at its greatest height, so its PE is maximum. (i) and (ii) are wrong: gravity (the ball's weight $mg$) still acts on it, so it has an acceleration $g$ downward; that is why it starts falling back.

(iii) and (iv)

See the table: e.g. truck uphill, chemical → kinetic + potential; photosynthesis, light → chemical; microphone, sound → electrical; solar panel, light → electrical.

(i) $$\text{Gain in PE}=mgh=50\times 10\times 72.5=36250\text{J}$$

(ii) The student reaches the same height, so the gain is again $mgh=36250$ J.

(iii) Potential energy depends only on the vertical height reached (the initial and final positions), not on the path taken.

(i) 36,250 J (ii) 36,250 J (iii) The gain in potential energy does not depend on the path, only on the change in height.

Let each floor be $h$ high and the first lift take time $t$.

• Energy to the 10th floor: $E_1=mg(10h)$. Energy to the 20th floor: $E_2=mg(20h)=2E_1$.
• Power: $P_1=\frac{E_1}{t}$, and $P_2=\frac{2E_1}{2t}=\frac{E_1}{t}=P_1$.

So twice the energy is needed (i.e. $E_1$ more), but the power is the same.

Energy required is double (one more E₁); the power required is the same as before.

• Factors: the energy needed is $mgh$, so it depends on the mass (weight) of the flag (and the part of the rope lifted with it), the height of the pole, and $g$. In practice, extra energy is needed to overcome friction in the pulley.
• Slowly or quickly: the work done is the same, because it depends only on the force (weight) and the height ($W=mgh$), not on the time taken.
• Doubling the speed: the same work is done in half the time, so the power $P=\frac{W}{t}$ is doubled.

It depends on the flag's mass, the height of the pole (and g, and pulley friction). The speed does not change the work done. Doubling the speed halves the time, so the power doubles.

The fuel used provides the kinetic energy $\frac{1}{2}M{v}^2$, where $M$ is the total moving mass.

• Day 1: $M_1=60+100=160$ kg
• Day 2: $M_2=60+100+40=200$ kg

$$\frac{{\text{Fuel}}_1}{{\text{Fuel}}_2}=\frac{\frac{1}{2}{M}_1{v}^2}{\frac{1}{2}{M}_2{v}^2}=\frac{160}{200}=\frac{4}{5}$$

4 : 5

A seesaw is a lever with the fulcrum in the middle. For balance, the turning effects on both sides must be equal:

$$W_{\text{child}}\times d_{\text{child}}=W_{\text{adult}}\times d_{\text{adult}}\Rightarrow W\times d_{\text{child}}=2W\times d\Rightarrow d_{\text{child}}=2d$$

So the child sits twice as far from the fulcrum as the adult.

The child sits at twice the distance of the adult from the fulcrum (child at 2d, adult at d), since W × 2d = 2W × d.

(i) Gravity acts downward. Upward motion: the displacement is opposite to gravity, so the work done by gravity is negative. Downward motion: the displacement is along gravity, so the work is positive.

(ii) Initial KE $=\frac{1}{2}\times 2\times 20^2=400$ J. At the top, KE = 0 and PE $=mgh=2\times 10\times 19.4=388$ J.

$$\text{Work by air resistance}=388-400=-12\text{J}$$

(Without air resistance the ball would have risen to $\frac{400}{20}=20$ m.)

(i) Negative while going up, positive while coming down. (ii) −12 J (12 J of energy lost to air resistance).

From Fig. 7.37 the force rises from 0 to 50 N between 0 and 1 m, stays at 50 N up to 3 m, and falls to 0 at 4 m. The work done is the area under the graph (a trapezium):

$$W=\frac{1}{2}(4+2)\times 50=150\text{J}$$

(i) At 0 m: $\frac{1}{2}m{v}_0^2=180\Rightarrow v_0^2=\frac{2\times 180}{10}=36\Rightarrow v_0=6$ m s⁻¹.

(ii) By the work–energy principle, KE at 4 m $=180+150=330$ J:

$$v^2=\frac{2\times 330}{10}=66\Rightarrow v=\sqrt{66}\approx 8.1{\text{m s}}^{-1}$$

No. The force always acts in the direction of motion (it is never negative), so the acceleration is positive or zero (zero only at 0 m and 4 m, where F = 0); the block keeps speeding up.

(i) 6 m s⁻¹ (ii) √66 ≈ 8.1 m s⁻¹. No negative acceleration: the force is never opposite to the motion.

At the highest point all the KE has become PE: $\frac{1}{2}m{u}^2=mgh\Rightarrow h=\frac{u^2}{2g}$. For the same $u$, $h\propto \frac{1}{g}$.

$$h_{\text{Moon}}=h_{\text{Earth}}\times \frac{g_{\text{Earth}}}{g_{\text{Moon}}}=8\times 6=48\text{m}$$

48 m

From Fig. 7.38: the speed is 35 m s⁻¹ from A (0 s) to B (1 s), then falls uniformly to zero at C (3 s).

(i) Between A and B the car moves with a constant speed of 35 m s⁻¹ (the driver's reaction time; the brakes have not yet been applied).

(ii) $$\text{KE at A}=\frac{1}{2}m{v}^2=\frac{1}{2}\times 1000\times 35^2=612500\text{J}$$

(iii) The brakes remove all the KE (work–energy principle):

$$W={\text{KE}}_C-{\text{KE}}_B=0-612500=-612500\text{J}$$

(negative, because the braking force acts opposite to the motion).

(iv) Into mainly heat (thermal energy) of the brakes, tyres and road, and a little sound. (The textbook answer key prints "potential energy" here, which is a slip: the car stays on a level road, so its potential energy does not change.)

(i) Constant speed of 35 m s⁻¹ (ii) 612,500 J (iii) −612,500 J (iv) Mainly heat (and some sound) produced by friction.

On a frictionless track the total mechanical energy stays the same: at O, KE = 0 and PE = 30 J, so total energy = 30 J everywhere. From Fig. 7.39, the PE is 20 J at P, 30 J at Q and 40 J at R.

• At P: KE $=30-20=10$ J. $\frac{1}{2}(0.5)v^2=10\Rightarrow v^2=40\Rightarrow v=2\sqrt{10}\approx 6.3$ m s⁻¹.
• At Q: KE $=30-30=0$, so $v=0$ m s⁻¹.
• At R: PE would have to be 40 J, which is more than the total energy of 30 J. So the ball can never reach R; it stops at Q and rolls back.

P: 2√10 ≈ 6.3 m s⁻¹; Q: 0 m s⁻¹; R: the ball cannot reach R (it would need 40 J but has only 30 J).

(i) PE lost = KE gained: $mgh=\frac{1}{2}m{v}^2$

$$v=\sqrt{2gh}=\sqrt{2\times 10\times 10}=\sqrt{200}=10\sqrt{2}\approx 14.1{\text{m s}}^{-1}\text{(downward)}$$

(ii) KE on impact $=mgh=1.5\times 10\times 10=150$ J. The sand does work $F\times d$ to stop it:

$$3000\times d=150\Rightarrow d=0.05\text{m}=5\text{cm}$$

(i) 10√2 ≈ 14.1 m s⁻¹ downward (ii) 0.05 m = 5 cm