Work, Energy and Power: JEE Advanced PYQ, 2007–2026

Answer: (C)

Pt = ½mv²: 16 × 4 = ½ × 2 × v², so v² = 64 and v = 8 m/s. (The force is not constant: it falls as the speed rises.)

Answer: (D)

Newton's second law gives T − mg = ma, so T = 2(10 + 2) = 24 N. Work by tension = T·d = 24 × 5 = 120 J. Check with the work-energy theorem: W_T + W_gravity = ΔKE, so 120 − 100 = 20 J, which equals ½mv² with v² = 2ad = 20.

Answer: (B)

Let θ be the angle of the radius to the block, measured from the vertical. The block has fallen a height R(1 − cosθ), so energy conservation gives v² = 2gR(1 − cosθ). Radially, mg cosθ − N = mv²/R. The block leaves when N = 0, so v² = gR cosθ. Equating: gR cosθ = 2gR(1 − cosθ), so 3cosθ = 2 and cosθ = 2/3. Then v² = gR·(2/3), so v = √(2gR/3).

Answer: (D)

Newton's second law for the lift (upward positive) gives T − mg = ma, so T = m(g + a) = 500 × 12 = 6000 N. The tension acts along the velocity, so the instantaneous power it delivers is P = Tv = 6000 × 3 = 18 000 W = 18 kW.

Answer: (B)

Power is the rate at which the pump does work, which equals the rate at which the water gains mechanical energy. Per second, 10 kg gains potential energy mgh = 10 × 10 × 5 = 500 J and kinetic energy ½mv² = ½ × 10 × 16 = 80 J. The total is 580 J per second, so P = 580 W. Taking only the potential term gives 500 W. Taking only the kinetic term gives 80 W. Using mv² in place of ½mv² gives 660 W.

Answer: (A)

The work-energy theorem gives W_gravity + W_drag = ΔKE. W_gravity = mgh = 0.002 × 10 × 500 = 10 J. ΔKE = ½(0.002)(6²) − 0 = 0.036 J. So W_drag = 0.036 − 10 = −9.964 J ≈ −9.96 J. Nearly all the gravitational work is dissipated by the air.

Answer: (A)

Use the work-energy theorem between the release point (at rest) and the lowest point (at rest). The net work is zero because the change in kinetic energy is zero. Gravity acts through the full drop (h + x) and does work mg(h + x). The spring force is variable and does work −½kx². So mg(h + x) − ½kx² = 0. Substituting m = 1 kg, g = 10 m/s², h = 0.60 m and k = 200 N/m gives 10(0.60 + x) = 100x², which simplifies to 100x² − 10x − 6 = 0. Then x = (10 + √(100 + 2400))/200 = (10 + 50)/200 = 0.30 m. Check: gravity does 10 × 0.90 = 9 J and the spring absorbs ½ × 200 × 0.09 = 9 J.

Answer: (A)

By the work-energy theorem, work equals the change in kinetic energy. W₁ = ½m[(2v)² − v²] = ½m(3v²) = 3mv²/2. W₂ = ½m[(3v)² − (2v)²] = ½m(5v²) = 5mv²/2. So W₂/W₁ = (5/2)/(3/2) = 5/3. Equal increments in speed do not need equal work, because kinetic energy depends on v².

Answer: (A)

The speed after each impact is e times the speed before it. The height reached is v²/2g, so each rebound height is e² times the previous drop height. The drops are h, e²h, e⁴h, … and the rises are e²h, e⁴h, … The total distance is D = h + 2(e²h + e⁴h + …) = h + 2h·e²/(1−e²) = h(1+e²)/(1−e²). With h = 1.8 m and e² = 0.25, D = 1.8 × 1.25/0.75 = 3.0 m.

Answer: (A)

Take the table top as the zero of PE. Initially a mass M/4 hangs with its centre of mass L/8 below the table, so U_i = −(M/4)g(L/8) = −MgL/32. At the end the whole chain hangs with its centre of mass L/2 below, so U_f = −MgL/2. The loss in PE is MgL(1/2 − 1/32) = 15MgL/32. This equals the KE gained: ½Mv² = 15MgL/32, so v² = 15gL/16 and v = ¼√(15gL).