Physics · Work, energy and power · chapter 5 of 14

Class 11 Physics Chapter 5: Work, Energy and Power case study question with answers

One case-based question of the kind CBSE-pattern school exams ask: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

A cement plant near Satna uses an inclined conveyor belt to load bags into a storage silo. Each bag has a mass of 50 kg. The belt carries 30 bags every minute and raises each bag through a vertical height of 6 m at a steady speed, so the bags gain no kinetic energy. The electric motor that drives the belt draws 4 kW from the supply. The plant engineer wants to know how much of this electrical power ends up as useful work on the bags, and what the motor costs to run. Take g = 10 m s⁻² and neglect the energy needed to start the belt.

Part 1 · 1 mark

The work done against gravity on the bags in one minute is

(a) 9 kJ(b) 1.5 kJ(c) 90 kJ(d) 900 kJ

Show answer

Answer: (c) 90 kJ

W = n mgh = 30 × 50 × 10 × 6 = 90 000 J = 90 kJ. 9 kJ is the work on three bags; 1.5 kJ is the work done in one second, not one minute.

Part 2 · 1 mark

Find the useful power delivered to the bags.

Answer and marking scheme

P = W/t = 90 000 J / 60 s = 1500 W = 1.5 kW.

P = 1500 W (1)

Part 3 · 2 marks

Find the efficiency of the conveyor system, and say where the rest of the energy goes.

Answer and marking scheme

Efficiency = useful power / input power = 1500/4000 = 0.375 = 37.5%. The remaining 2.5 kW is lost as heat in the motor windings and through friction in the rollers and belt, with some sound.

efficiency = 37.5% (1)
rest lost as heat and friction in motor and belt (1)

The belt runs for an 8-hour shift. Find the electrical energy the motor uses in kWh, and the cost at ₹7 per kWh.

Answer and marking scheme

Energy = 4 kW × 8 h = 32 kWh. Cost = 32 × ₹7 = ₹224.

32 kWh (1)
₹224 (1)

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