Case Study 1: Definition of Work
Ravi applied a force of 20 N on a box and pushed it 5 m along the floor in the direction of the force. Work done is the product of force and displacement in the direction of the force.
Questions:
- Work = ?
(a) F + d (b) F/d (c) F × d (d) F × v
Answer: (c) - Work done = 20 × 5 = ?
(a) 50 J (b) 100 J (c) 200 J (d) 500 J
Answer: (c) - SI unit of work = ?
(a) Newton (b) Joule (c) Watt (d) Pascal
Answer: (b) - Work is done only if:
(a) force is applied (b) displacement occurs (c) force has a component in displacement direction (d) all
Answer: (d) - If displacement is zero, work = ?
(a) zero (b) infinite (c) constant (d) negative
Answer: (a)
Case Study 2: Work Done Against Friction
Aman pushes a box of mass 10 kg on the floor, applying 50 N force, but 20 N is lost against friction.
Questions:
- Net force used for displacement = ?
(a) 20 N (b) 30 N (c) 50 N (d) 70 N
Answer: (b) - If displacement = 4 m, work done = ?
(a) 80 J (b) 100 J (c) 120 J (d) 200 J
Answer: (c) - Work done against friction = ?
(a) 20 J (b) 40 J (c) 60 J (d) 80 J
Answer: (d) - Type of work against friction = ?
(a) positive (b) negative (c) zero (d) none
Answer: (b) - Work done depends on:
(a) force only (b) displacement only (c) angle between force & displacement (d) all
Answer: (d)
Case Study 3: Positive and Negative Work
A man lifts a load upward against gravity (positive work). When he lowers it, gravity does work on it (positive) while his force does negative work.
Questions:
- Positive work is done when:
(a) force and displacement are same direction (b) opposite (c) perpendicular (d) none
Answer: (a) - Negative work is done when:
(a) force opposes displacement (b) both same (c) both zero (d) none
Answer: (a) - Work done by gravity in free fall = ?
(a) positive (b) negative (c) zero (d) none
Answer: (a) - Work done by a man lowering object = ?
(a) positive (b) negative (c) zero (d) none
Answer: (b) - If force is perpendicular to displacement (e.g., circular motion), work = ?
(a) 0 (b) positive (c) negative (d) none
Answer: (a)
Case Study 4: Kinetic Energy
A car of mass 1000 kg moves at 10 m/s. Its kinetic energy is ½ mv².
Questions:
- KE = ?
(a) ½ mv² (b) mv (c) mgh (d) none
Answer: (a) - KE = ½ × 1000 × (10)² = ?
(a) 5,000 J (b) 10,000 J (c) 50,000 J (d) 100,000 J
Answer: (c) - KE depends on:
(a) mass only (b) velocity only (c) both (d) none
Answer: (c) - If velocity doubled, KE becomes:
(a) 2 times (b) 3 times (c) 4 times (d) 8 times
Answer: (c) - KE is a form of:
(a) potential energy (b) mechanical energy (c) chemical energy (d) none
Answer: (b)
Case Study 5: Potential Energy
A stone of mass 2 kg is lifted to a height of 5 m. PE = mgh.
Questions:
- PE = ?
(a) mgh (b) ½ mv² (c) force × displacement (d) none
Answer: (a) - Value = 2 × 9.8 × 5 = ?
(a) 49 J (b) 98 J (c) 100 J (d) 196 J
Answer: (d) - PE depends on:
(a) mass only (b) velocity (c) height (d) both mass & height
Answer: (d) - SI unit of PE = ?
(a) Joule (b) Watt (c) Newton (d) Pascal
Answer: (a) - If height doubled, PE becomes:
(a) double (b) triple (c) half (d) none
Answer: (a)
Case Study 6: Work–Energy Theorem
A moving car stops after brakes are applied. Its kinetic energy is converted into heat and sound. Work done by brakes = loss in KE.
Questions:
- Work–energy theorem:
(a) Work = change in KE (b) Work = KE only (c) Work = PE only (d) none
Answer: (a) - Brakes apply:
(a) positive work (b) negative work (c) zero (d) none
Answer: (b) - Energy conversion:
(a) KE → PE (b) KE → heat & sound (c) PE → KE (d) none
Answer: (b) - If KE = 2000 J and car stops, work done by brakes = ?
(a) 0 (b) –2000 J (c) +2000 J (d) none
Answer: (b) - Work–energy theorem applies to:
(a) uniform motion only (b) accelerated motion only (c) all motions (d) none
Answer: (c)
Case Study 7: Power
A boy lifts a 20 kg load to 2 m in 10 seconds. Power = work/time.
Questions:
- Work done = mgh = ?
(a) 200 J (b) 300 J (c) 400 J (d) none
Answer: (c) - Power = 400/10 = ?
(a) 20 W (b) 30 W (c) 40 W (d) 50 W
Answer: (c) - SI unit of power = ?
(a) joule (b) watt (c) newton (d) pascal
Answer: (b) - 1 watt = work done of 1 joule in:
(a) 1 sec (b) 10 sec (c) 100 sec (d) none
Answer: (a) - Power depends on:
(a) work only (b) time only (c) both (d) none
Answer: (c)
Case Study 8: Commercial Unit of Energy
Electricity bills are measured in kilowatt-hour (kWh). 1 kWh = 1 unit.
Questions:
- 1 kWh = ?
(a) 1000 J (b) 1000 W × 1 s (c) 1000 W × 3600 s (d) none
Answer: (c) - 1 kWh in joules = ?
(a) 3.6 × 10³ J (b) 3.6 × 10⁵ J (c) 3.6 × 10⁶ J (d) 3.6 × 10⁶ J
Answer: (d) - SI unit of energy = ?
(a) watt (b) joule (c) kWh (d) pascal
Answer: (b) - Commercial unit of energy = ?
(a) joule (b) newton (c) kWh (d) none
Answer: (c) - If 2 kW heater runs for 3 hours, energy = ?
(a) 6 kWh (b) 3 kWh (c) 2 kWh (d) none
Answer: (a)
Case Study 9: Transformation of Energy
In a pendulum, energy transforms from PE at highest point to KE at lowest point, and back again.
Questions:
- At highest point, pendulum has:
(a) KE only (b) PE only (c) both (d) none
Answer: (b) - At mean position, pendulum has:
(a) KE only (b) PE only (c) both equal (d) none
Answer: (a) - Total energy remains:
(a) increasing (b) decreasing (c) constant (d) zero
Answer: (c) - This is an example of:
(a) conservation of energy (b) conservation of momentum (c) work–energy theorem (d) none
Answer: (a) - Energy lost in real pendulum due to:
(a) gravity (b) air resistance (c) tension (d) none
Answer: (b)
Case Study 10: Work in Daily Life
A coolie carrying a load on his head and walking on a horizontal road does no work against gravity, but he does work against friction.
Questions:
- Work against gravity in horizontal motion = ?
(a) positive (b) negative (c) zero (d) none
Answer: (c) - Work against friction = ?
(a) positive (b) negative (c) zero (d) none
Answer: (a) - If no displacement, work = ?
(a) zero (b) positive (c) negative (d) none
Answer: (a) - Work is scalar or vector?
(a) scalar (b) vector (c) both (d) none
Answer: (a) - In physics, “work” depends on:
(a) physical effort (b) displacement in direction of force (c) only energy (d) none
Answer: (b)