Think It Over, Activities & Pause-and-Ponder
About a child sliding down the playground slides:
(a) Velocity at the bottom of the blue slide: Using conservation of mechanical energy, the potential energy at the top converts to kinetic energy at the bottom:
(b) Two children of different masses: Since $v=\sqrt{2gh}$ does not contain mass, both children reach the bottom with the same velocity (ignoring friction).
(c) Which slide gives the largest velocity: The velocity depends only on the height $h$, not on the shape or length of the slide. So the slide with the greatest height gives the largest velocity at the bottom.
Key conclusions from the chapter’s activities.
- Activity 7.1 (ball in sand): A ball dropped from a greater height makes a deeper depression. Greater height → more potential energy ($U=mgh$).
- Activity 7.2 (pendulum): The bob rises to almost the same height on the other side, showing mechanical energy is conserved (PE ↔ KE). It eventually stops due to friction and air resistance.
- Activity 7.3 (inclined plank): As the plank becomes less steep, the force needed to pull the cart up decreases, but it must be applied over a longer distance.
- Activity 7.4 (scale lever): A light eraser can lift a heavier stapler because the scale acts as a lever — a small effort over a long arm balances a large load on a short arm.
- Activity 7.5 (beam balance): The beam balances when $n_1 L_1 = n_2 L_2$, i.e., effort × effort arm = load × load arm.
1. A weightlifter holds a barbell steady in her hands. Is she doing any work on the barbell while holding it steady?
No. Work done $W = F \times s$. While holding the barbell steady, the displacement $s = 0$, so the work done on the barbell is zero. She still feels tired because her muscles use up internal (chemical) energy to keep applying the force, but in the scientific sense no work is done.
2. Is the work done by friction on a stack of coins moving on a rough surface positive, negative or zero?
Negative. Friction always acts opposite to the direction of motion (displacement). Since force and displacement are in opposite directions, the work done by friction on the moving coins is negative.
3. When you pedal a bicycle on a flat road, your muscles supply energy. In what forms does this muscular energy appear?
The chemical (muscular) energy is converted mainly into:
- Kinetic energy of the moving bicycle and rider.
- Heat (thermal) energy due to friction in the body, the chain, the tyres and the air.
- A small amount of sound energy (the rolling and mechanical noises).
On a flat road the height doesn’t change, so there is no gain in gravitational potential energy.
4. Two objects A and B of mass $m$ and $4m$ have the same kinetic energy. Find the ratio of the magnitudes of their velocities.
5. Does the kinetic energy of an object moving with constant velocity change with its position?
No. Kinetic energy $K=\tfrac{1}{2}mv^2$ depends only on the object’s speed, not on its position. If the velocity is constant, the kinetic energy stays the same at every position.
6. Does the potential energy of an object near the Earth change if it moves with constant velocity horizontally? What if it is gradually raised vertically?
Gravitational potential energy $U = mgh$ depends only on the height $h$.
- Horizontal motion: the height does not change, so the potential energy does not change.
- Raised vertically: the height increases, so the potential energy increases.
7. For the freely falling ball (Fig. 7.19), calculate the mechanical energy just before it hits the ground and show it equals $mgh$.
Just before hitting the ground, the height is $0$, so potential energy $= 0$. Using $v^2 = 2gh$ (object dropped from height $h$):
So even at the ground the mechanical energy is $mgh$ — equal to its value at the top. This confirms the conservation of mechanical energy.
8. A ball is released from the highest point of a science-park track (Fig. 7.22). Describe how KE and PE change at A, B and C. Why are later points (C, D, E) lower? Is friction involved?
- At A (highest point): the ball is momentarily slow/at rest, so PE is maximum and KE is minimum.
- Rolling down to B (a low point): PE converts into KE, so KE is maximum and PE is minimum at B.
- Rising to C: KE converts back into PE, so KE decreases and PE increases again.
Each successive peak (C, D, E) is lower than the previous one because some mechanical energy is continuously lost as heat (and sound) due to friction and air resistance. With less total mechanical energy left, the ball cannot rise as high. Yes — it is because of energy lost to friction.
9. Why are roads on hills built to wind around in gentle slopes rather than going straight up?
A winding road is a long inclined plane with a small slope, so its length $L$ is large compared to the height $h$. Mechanical advantage $=\dfrac{L}{h}$, so a larger $L$ means a greater mechanical advantage and much less force is needed to climb. A straight, steep climb would need a very large force, which vehicles could not easily provide. The trade-off is that the vehicle travels a longer distance.
10. Why is climbing an inclined ladder easier than climbing a vertical ladder?
An inclined ladder acts as an inclined plane with mechanical advantage $\dfrac{L}{h} > 1$, so less force/effort is needed per step to gain height. A vertical ladder has $L = h$ (mechanical advantage $= 1$), so you must support nearly your full weight. The inclined path needs less effort, though you cover a longer distance.
11. Why is it easier to open the lid of a can using a spoon (Fig. 7.35)?
The spoon works as a lever (Class I). The rim of the can acts as the fulcrum. The effort arm (the long handle of the spoon) is much longer than the load arm (short distance from the rim to the lid). So mechanical advantage $=\dfrac{\text{effort arm}}{\text{load arm}} > 1$, and a small effort produces a large force on the lid, prying it open easily.
12. Why do you push a hard object closer to the scissors’ fulcrum when cutting it?
Placing the object close to the fulcrum makes the load arm short. For the same effort applied at the long handles, the force on the object equals $\text{effort}\times\dfrac{\text{effort arm}}{\text{load arm}}$. A shorter load arm means a larger cutting force, which is needed to cut a hard object.
13. Why do all real machines eventually slow down and stop (no perpetual machine works)? Explain in terms of work and energy.
In every real machine, friction and air resistance continuously do negative work, converting part of the useful mechanical energy into heat and sound that cannot be recovered as work. Because energy keeps leaking away, the total mechanical energy steadily decreases, so the machine gradually slows down and stops. A perpetual machine is impossible because it would have to keep doing work without any energy input, which violates the principle that energy cannot be created — and some is always lost to friction.
Revise, Reflect, Refine — Exercises
1. State True or False.
(i) Work is done when a force is applied, even if the object does not move. → False. If $s=0$, then $W = F\times 0 = 0$; displacement is needed for work.
(ii) Lifting a bucket vertically upward does positive work on it. → True. The applied force and displacement are both upward (same direction).
(iii) The SI unit for both work and energy is the joule (J). → True.
(iv) A motionless stretched rubber band has kinetic energy. → False. It is not moving, so it has potential (elastic) energy, not kinetic energy.
(v) Energy can change from one form to another. → True.
2. Fill in the blanks.
- (i) Work done = force × displacement (in the direction of force).
- (ii) 1 joule of work is done when a force of 1 newton displaces an object by 1 metre.
- (iii) Kinetic energy of a body of mass $m$ and velocity $v$ is $\boldsymbol{\tfrac{1}{2}mv^2}$.
- (iv) Potential energy of mass $m$ at height $h$ is $\boldsymbol{mgh}$.
- (v) Power is defined as the rate at which work is done.
3. When a ball thrown upward reaches its highest point, which statements are correct?
At the highest point the velocity is zero, so KE = 0 (iii ✓) and the height is greatest, so PE is maximum (iv ✓). However, gravity still acts on the ball, so the force is $mg \neq 0$ (i ✗) and the acceleration is $g \neq 0$ (ii ✗).
4. Identify the energy transformation in each situation.
| Situation | Energy transformation |
|---|---|
| (i) Truck moving uphill | Chemical (fuel) → Kinetic + Potential |
| (ii) Unwinding of a watch spring | Elastic potential → Kinetic (mechanical) |
| (iii) Photosynthesis in green leaves | Light (solar) → Chemical |
| (iv) Water flowing from a dam | Potential → Kinetic |
| (v) Burning of a matchstick | Chemical → Heat + Light |
| (vi) Explosion of a firecracker | Chemical → Heat + Light + Sound + Kinetic |
| (vii) Speaking into a microphone | Sound → Electrical |
| (viii) Glowing electric bulb | Electrical → Light + Heat |
| (ix) Solar panel | Light (solar) → Electrical |
5. A student ($m=50$ kg) is taken to the top of a building of height $h=72.5$ m, first by elevator and then by stairs. ($g=10\,\text{m s}^{-2}$.)
(i) Lifted straight up by elevator:
(ii) Climbing the stairs to the same top: the height gained is still $72.5$ m, so
(iii) Conclusion: Gravitational potential energy depends only on the vertical height, not on the path taken.
6. A crane lifts mass $m$ to the 10th floor in time $t$. Then it raises the same mass to the 20th floor in time $2t$. How much more energy and power are required? (Equal floor heights.)
Let the height of one floor be $h_0$. Reaching the 20th floor means twice the height of the 10th floor.
So twice the energy is needed (height doubled), but since the time is also doubled, the power is unchanged.
7. Which factors decide the energy to raise a flag up a flagpole using a pulley? Does raising it slowly or quickly change the work? If the speed is doubled, how does the power change?
The energy (work) needed is $W = mgh$, which depends on the mass of the flag $m$, the height of the flagpole $h$, and $g$. (A fixed pulley only changes the direction of the effort, not the work.)
Slow or quick? The work done is the same either way, because $W = mgh$ does not depend on time or speed.
Doubling the speed: the same work is done in half the time. Since power $P = \dfrac{W}{t}$, halving $t$ doubles the power.
8. Day 1: man (60 kg) + scooter (100 kg) reach speed $v$. Day 2: man + son (40 kg) + scooter reach the same $v$ in the same time. Find the ratio of fuel used on the two days. (Energy goes entirely to the scooter; ignore other losses.)
Fuel used is proportional to the kinetic energy gained, $K=\tfrac{1}{2}Mv^2$.
9. On a seesaw, a child and an adult (twice the child’s weight) balance. Draw a figure showing their distances from the fulcrum.
For balance: $\text{load}\times\text{load arm} = \text{effort}\times\text{effort arm}$. Let the child’s weight be $W$ and the adult’s be $2W$:
So the lighter child must sit twice as far from the fulcrum as the heavier adult.
10. A 2 kg ball is thrown up at $20\,\text{m s}^{-1}$. ($g=10\,\text{m s}^{-2}$.)
Upward motion: gravity acts downward, displacement is upward → opposite directions → negative work.
Downward motion: gravity acts downward, displacement is downward → same direction → positive work.
(Without air resistance the ball would have reached $20$ m; the missing $0.6$ m of height corresponds to the 12 J lost.)
11. A 10.0 kg block moves on a frictionless floor. A variable force acts in the direction of motion from 0 m to 4 m (graph below). The block has KE = 180 J at 0 m. Find its speed at (i) 0 m and (ii) 4 m. Is there any negative acceleration?
(i) Speed at 0 m:
Work done by the force = area under the graph (a trapezium):
(ii) Speed at 4 m: By the work–energy theorem, KE at 4 m $= 180 + 150 = 330\ \text{J}$.
Negative acceleration? No. The force always acts in the direction of motion (it is positive throughout, even while decreasing from 3 m to 4 m). So acceleration $a = F/m$ stays positive, and the block keeps speeding up the whole way.
12. Moon’s gravity is $\tfrac{1}{6}$ of Earth’s. A ball thrown up reaches 8 m on Earth. How high will the same throw go on the Moon?
Maximum height $h = \dfrac{v^2}{2g}$. For the same throwing velocity $v$, $h \propto \dfrac{1}{g}$.
13. A 1000 kg car (speed–time graph below) moves, then brakes to a stop.
(i) Between A and B: the speed is constant at 35 m s⁻¹ — this is the driver’s reaction time, before the brakes are applied.
(ii) Kinetic energy at A:
(iii) Work done by brakes (B to C): the car stops, so all the kinetic energy is removed.
Work by brakes = − 612500 J (≈ −6.13 × 10⁵ J)(iv) The kinetic energy transforms mainly into heat (thermal energy) in the brakes (and a little sound).
14. A 0.5 kg ball on a frictionless track has PE–displacement graph below. At O, $v=0$ and PE $=30$ J. Find the velocity at P, Q and R.
The track is frictionless, so the total mechanical energy is conserved. At O: KE $=0$, PE $=30$ J, so ME $= 30$ J everywhere. At any point, KE $= 30 – \text{PE}$, and $v=\sqrt{\dfrac{2K}{m}}$ with $m = 0.5$ kg.
At P (PE $= 20$ J): $K = 30-20 = 10$ J
At Q (PE $= 30$ J): $K = 30-30 = 0$ J
At R (PE $= 40$ J): $K = 30-40 = -10$ J, which is impossible.
15. A 1.5 kg coconut falls from a 10 m tree onto wet sand and stops, making a depression. ($g=10\,\text{m s}^{-2}$.)
The coconut’s energy on impact = its PE at the top = $mgh$. This is used up by the resistive force $F = 3000$ N over depth $d$ (Work = $F\times d$).