Work and Energy — Practice

Question 1

Define work and state its SI unit.

Accepted Answer

Step 1 — Define.
  Work is done when a force acts on a body and produces a displacement in the direction of the force.
  Symbolic: W = F × s.

Step 2 — SI unit.
  joule (J), where 1 J = 1 N × 1 m = 1 kg·m²/s².

✦ Answer: Work = force × displacement (in the direction of the force). SI unit: joule.

Question 2

Find the kinetic energy of a 2 kg object moving at 10 m/s.

Accepted Answer

Step 1 — KE = ½mv².
  = ½ × 2 × 10² = ½ × 2 × 100 = 100 J.

✦ Answer: 100 J.

Question 3

A 5 kg object is raised to a height of 3 m. Find its potential energy. (g = 10 m/s²)

Accepted Answer

Step 1 — PE = mgh = 5 × 10 × 3 = 150 J.

✦ Answer: 150 J.

Question 4

Give two examples where the force applied on an object does zero work.

Accepted Answer

Step 1 — Recall: work = 0 if (i) no displacement OR (ii) force perpendicular to displacement.

Step 2 — Examples.

(i) Pushing a wall that doesn't move:
  Force applied, but no displacement → W = 0.

(ii) Carrying a suitcase while walking on a horizontal floor:
  Lifting force is vertical; motion is horizontal → force perpendicular to displacement → W = 0 on the suitcase.

(Alternative) (iii) Satellite orbiting in a perfect circle: gravity points to centre (perpendicular to motion) → W = 0 by gravity. Energy doesn't change → orbit speed constant.

✦ Answer: (i) Pushing a wall (no displacement); (ii) Carrying a suitcase horizontally (force perpendicular to motion).

Question 5

A man lifts a load of 20 kg through a height of 5 m in 10 s. Find the work done and his power. (g = 10 m/s²)

Accepted Answer

Step 1 — Find work.
  W = F × s = mg × h = 20 × 10 × 5 = 1000 J.

Step 2 — Find power.
  P = W / t = 1000 / 10 = 100 W.

✦ Answer: Work = 1000 J. Power = 100 W.

A typical adult can sustain ~ 100 W of mechanical power over an hour or two. Athletes peak at 500–1500 W for short bursts.

Question 6

A 2 kg ball moving at 5 m/s is brought to rest by a force of 20 N. How far does it travel before stopping?

Accepted Answer

Step 1 — Use work-energy theorem.
  W = ΔKE.
  Initial KE = ½ × 2 × 5² = 25 J. Final KE = 0.
  W = 0 − 25 = −25 J (negative because the force decelerates the body).

Step 2 — Relate to F × s.
  Braking force F = 20 N (opposite to motion).
  W = −F × s ⇒ −25 = −20 × s ⇒ s = 1.25 m.

✦ Answer: 1.25 m.

Work-energy theorem shortcut: avoided calculating acceleration explicitly.

Question 7

A ball of mass 0.5 kg is dropped from a height of 10 m. Find its kinetic energy and velocity just before it hits the ground. (g = 10 m/s²)

Accepted Answer

Step 1 — Use conservation of mechanical energy.
  At top: PE = mgh = 0.5 × 10 × 10 = 50 J. KE = 0.
  At bottom: PE = 0. KE = total energy.
  KE_bottom = 50 J.

Step 2 — Find velocity.
  KE = ½mv² = 50 ⇒ ½ × 0.5 × v² = 50 ⇒ v² = 200 ⇒ v ≈ 14.14 m/s.

✦ Answer: KE just before impact = 50 J. Velocity ≈ 14.14 m/s.

Verification with v² = 2gh: 2 × 10 × 10 = 200 ⇒ v ≈ 14.14 m/s. ✓

Question 8

An electric bulb of 100 W is used for 10 hours daily. Calculate the energy consumed in 30 days in (a) joules, (b) kilowatt-hours.

Accepted Answer

Step 1 — Total time in 30 days.
  t = 30 × 10 = 300 hours.

Step 2 — Energy in kWh.
  E = P × t = 0.1 kW × 300 h = 30 kWh.

Step 3 — Energy in joules.
  1 kWh = 3.6 × 10⁶ J.
  E = 30 × 3.6 × 10⁶ = 1.08 × 10⁸ J.

✦ Answer: (a) 1.08 × 10⁸ J. (b) 30 kWh = 30 'units'.

Question 9

A ball thrown upward decelerates from 20 m/s to 0 m/s while rising a height of 20 m. Calculate (a) the work done by gravity on the ball, (b) the change in its kinetic energy. (m = 1 kg, g = 10 m/s²)

Accepted Answer

Step 1 (a) — Force of gravity is downward; displacement is upward (opposite directions).
  W_gravity = −F × s = −mg × h = −1 × 10 × 20 = −200 J.

Step 2 (b) — Change in KE.
  Initial KE = ½ × 1 × 20² = 200 J.
  Final KE = 0.
  ΔKE = 0 − 200 = −200 J.

Step 3 — Note that W = ΔKE = −200 J. The work-energy theorem holds: gravity took 200 J of KE from the ball (which went into PE).

✦ Answer: (a) W_gravity = −200 J. (b) ΔKE = −200 J. They are equal — exactly as predicted by the work-energy theorem.

Question 10

Two bodies A and B of masses 1 kg and 4 kg have the same kinetic energy. Find the ratio of their velocities.

Accepted Answer

Step 1 — Same KE.
  ½ m_A v_A² = ½ m_B v_B²
  m_A × v_A² = m_B × v_B².

Step 2 — Substitute.
  1 × v_A² = 4 × v_B²
  v_A² / v_B² = 4
  v_A / v_B = 2.

✦ Answer: v_A : v_B = 2 : 1.

The lighter body needs to move FASTER to have the same KE.

Question 11

A 100 g cricket ball moving horizontally at 30 m/s is brought to rest by a fielder's hands in 0.1 s. Calculate (a) the average force exerted by the fielder's hands, (b) the work done by the hands.

Accepted Answer

Step 1 — Convert mass.
  m = 100 g = 0.1 kg.

Step 2 (a) — Use impulse-momentum.
  F × t = m × Δv ⇒ F × 0.1 = 0.1 × 30 = 3 ⇒ F = 30 N.

Step 3 (b) — Use work-energy theorem.
  Initial KE = ½ × 0.1 × 30² = ½ × 0.1 × 900 = 45 J.
  Final KE = 0.
  W_hands = ΔKE = 0 − 45 = −45 J. The hands do −45 J of work on the ball (taking 45 J of KE away). Equivalently, the BALL does +45 J of work on the hands.

✦ Answer: (a) Force = 30 N. (b) Work done by hands on ball = −45 J (negative; hands absorbed 45 J of KE).

Question 12

A stone of mass 2 kg is thrown vertically upward with an initial velocity of 20 m/s. Find (a) the maximum height reached, (b) the total energy at any point during the motion. (g = 10 m/s²)

Accepted Answer

Step 1 (a) — Use v² = u² − 2gh at top (v = 0).
  0 = 400 − 20h ⇒ h = 20 m.

Step 2 (b) — Total mechanical energy.
  At the start (h = 0): KE = ½ × 2 × 20² = 400 J. PE = 0. Total = 400 J.
  At the top (h = 20): KE = 0. PE = mgh = 2 × 10 × 20 = 400 J. Total = 400 J. ✓
  At any intermediate point: KE + PE = 400 J (no air resistance).

Step 3 — Note conservation.
  As the stone rises, KE → PE. As it falls, PE → KE. Total never changes (frictionless).

✦ Answer: (a) 20 m; (b) 400 J — constant throughout the motion.

Question 13

A household uses 5 bulbs of 60 W each for 5 hours a day, 1 fan of 75 W for 8 hours a day, and a fridge of 100 W running continuously. Calculate the daily and monthly (30 days) electricity consumption in kWh, and the cost at Rs. 6 per unit.

Accepted Answer

Step 1 — Bulbs.
  5 × 60 = 300 W × 5 h = 1500 Wh = 1.5 kWh per day.

Step 2 — Fan.
  75 W × 8 h = 600 Wh = 0.6 kWh per day.

Step 3 — Fridge.
  100 W × 24 h = 2400 Wh = 2.4 kWh per day.

Step 4 — Daily total.
  1.5 + 0.6 + 2.4 = 4.5 kWh/day.

Step 5 — Monthly total.
  4.5 × 30 = 135 kWh.

Step 6 — Cost.
  135 × 6 = Rs. 810.

✦ Answer: Daily consumption = 4.5 kWh. Monthly = 135 kWh. Monthly bill = Rs. 810.

Question 14

An object of mass 40 kg is raised through a height of 5 m on the moon. Compare the work done with that on Earth. Also, when the object is dropped from this height, find its velocity just before hitting the ground (a) on Earth, (b) on Moon. (g_E = 10 m/s², g_M = 1.6 m/s²)

Accepted Answer

Step 1 — Work done to lift an object = mgh.
  On Earth: W_E = 40 × 10 × 5 = 2000 J.
  On Moon: W_M = 40 × 1.6 × 5 = 320 J.
  Ratio: W_E / W_M = 10/1.6 ≈ 6.25. Lifting on Moon takes only 1/6.25 = 16 % of the Earth's work.

Step 2 — Velocity on dropping.
  PE at height = KE at ground.
  ½ m v² = mgh ⇒ v = √(2gh).
  (a) Earth: v = √(2 × 10 × 5) = √100 = 10 m/s.
  (b) Moon: v = √(2 × 1.6 × 5) = √16 = 4 m/s.

Step 3 — Why same conclusion?
  Less work to lift on Moon → less PE → less KE on landing → smaller v.

✦ Answer:
  Lifting work: Earth 2000 J, Moon 320 J (factor 6.25).
  Landing velocity: Earth 10 m/s, Moon 4 m/s (factor 2.5).

Question 15

Why is the kinetic energy of a body always positive while its potential energy can be positive, negative or zero?

Accepted Answer

Step 1 — KE formula. KE = ½mv². Mass is always positive. v² is the square of velocity, always ≥ 0. Hence KE ≥ 0 always. KE can be zero (body at rest) or positive (body moving); never negative. Step 2 — PE formula. PE = mgh. m and g are positive. h is height ABOVE the chosen reference level. • h > 0 (above reference): PE > 0. • h = 0 (at reference): PE = 0. • h < 0 (below reference): PE < 0 (mathematically allowed; physically real — a basement is at negative PE relative to the ground floor). Step 3 — Practical example. A miner at 100 m below ground has PE = −100 mg J relative to the surface. Negative because you need to do positive work AGAINST gravity to bring her back up. ✦ Answer: KE is always positive because it depends on v² (always ≥ 0). PE depends on height ABOVE the reference; can be negative if the body is below the reference. The reference is arbitrary, and so is the sign — only DIFFERENCES in PE matter physically.

Practice — Work and Energy

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