Class 11 Physics Chapter 5: Work, Energy and Power worksheet with answer key

Answer: (d) 14/3

Component of A⃗ along B⃗ = A⃗·B⃗ / B = (8 − 2 + 8)/√(4 + 1 + 4) = 14/3. Not dividing by B gives 14; dividing by B² = 9 gives 14/9, and by A = 6 gives 7/3.

Answer: (d) 25 J

W = Fd cos θ = 10 × 5 × cos 60° = 25 J. Using sin 60° gives 43.3 J; leaving out the angle gives 50 J.

Answer: (b) −12

A⃗·B⃗ = AB cos θ = 4 × 6 × cos 120° = 24 × (−½) = −12. Taking cos 120° as +½ gives 12; leaving out cos θ gives 24.

Answer: (a) [MT⁻²]

k = force/extension = [MLT⁻²]/[L] = [MT⁻²]; its unit is N m⁻¹. [MLT⁻²] is force and [ML²T⁻²] is energy.

Answer: (c) 45 kWh

Energy = power × time = 2 kW × 0.75 h × 30 = 45 kWh. Using 45 h instead of 0.75 h each day gives 2700 kWh; using 1 h each day gives 60 kWh.

(i) V = mgh = 2 × 10 × 1.5 = 30 J. (ii) V = 2 × 10 × (1.5 − 0.8) = 14 J. Shelf to table: with the floor as zero, ΔV = 2 × 10 × 0.8 − 30 = 16 − 30 = −14 J; with the table top as zero, ΔV = 0 − 14 = −14 J. The zero is arbitrary; only changes in potential energy matter.

• 30 J and 14 J (1)
• ΔV = −14 J in both cases (1)

Zero. The tension acts along the string, towards the centre, while the stone's displacement at every instant is along the tangent; θ = 90°, so the tension does no work. The speed is constant, so the kinetic energy does not change, ΔK = 0, which the work-energy theorem requires for zero net work.

• work is zero, tension perpendicular to the displacement (1)
• ΔK = 0 at constant speed, consistent with the theorem (1)

Energy each day = 1.5 kW × (40/60) h = 1.0 kWh. In 30 days: 1.0 × 30 = 30 kWh. Cost = 30 × ₹8 = ₹240.

• energy = 30 kWh (1)
• cost = ₹240 (1)

K = ½mv², so dK/dt = mv (dv/dt) = (m dv/dt) v. By Newton's second law m dv/dt = F, so dK/dt = Fv = F dx/dt, giving dK = F dx. Integrating from x_i to x_f: K_f − K_i = ∫ F dx (from x_i to x_f). The right side is, by definition, the work W done by the variable force, so K_f − K_i = W: the change in kinetic energy equals the work done by the net force.

• dK/dt = mv dv/dt = Fv (1)
• dK = F dx (1)
• integration gives K_f − K_i = ∫F dx = W (1)

(a) At the maximum speed the engine force balances the resistance: P = Fv, v = 30000/1500 = 20 m s⁻¹. (b) Up the slope the engine must also balance mg sin θ = 1200 × 10 × (1/20) = 600 N, so F = 1500 + 600 = 2100 N and v = 30000/2100 = 14.3 m s⁻¹.

• v = P/F = 20 m s⁻¹ on the level (1)
• extra force mg sin θ = 600 N (1)
• v = 14.3 m s⁻¹ up the slope (1)

(a) E = ½kx_m² = ½ × 80 × (0.05)² = 0.1 J. (b) At x = 0 all of it is kinetic: ½mv_m² = 0.1 J, v_m = √(2 × 0.1/0.2) = 1 m s⁻¹. (c) K = V means V = E/2: ½kx² = ½ × ½kx_m², so x = x_m/√2 = 5/1.414 = 3.54 cm on either side of the equilibrium position.

• E = 0.1 J (1)
• v_m = 1 m s⁻¹ (1)
• x = ±3.54 cm (1)