Questions & explanations
1. A simple pendulum of length 2.0 m with a bob of mass 0.50 kg is released from rest at an angle of 30° from vertical. Air drag dissipates energy at a constant average power of 0.10 W. What is the approximate speed of the bob at the lowest point after it has completed one half-cycle (from release to the lowest point)?
- 1.7 m/s
- 2.0 m/s
- 2.6 m/s
- 2.3 m/s
Answer: 2.3 m/s
Using energy conservation, initial height h = L(1 - cos30°) = 2.0(1 - 0.8660) = 0.268 m. Initial mechanical energy E0 = mgh = 0.50 × 10 × 0.268 = 1.34 J. Period T = 2π√(L/g) ≈ 2.84 s, half-period = 1.42 s. Energy lost to drag = P × (T/2) = 0.10 × 1.42 = 0.142 J. Remaining energy at lowest point = 1.34 - 0.142 = 1.198 J, all kinetic. Speed v = √(2E/m) = √(2 × 1.198 / 0.50) = √4.792 ≈ 2.19 m/s, which rounds to 2.3 m/s.
2. A 1 kg block slides on a rough horizontal surface (μ = 0.4) with initial speed 6 m/s. It then compresses a spring (k = 100 N/m). If the block travels 2 m before hitting the spring, find the maximum compression of the spring. (g = 10 m/s²)
- 0.8 m
- 0.2 m
- 0.6 m
- 0.4 m
Answer: 0.4 m
Initial KE = (1/2)×1×36 = 18 J. Work done against friction before hitting spring = μ mg d = 0.4×1×10×2 = 8 J. Energy left = 10 J. At max compression, spring PE = (1/2)kx² = 50x². Also friction during compression does work = μ mg x = 4x. Energy balance: 18 = 8 + 4x + 50x² => 50x² + 4x - 10 = 0 => x = 0.4 m (positive root).
3. Two blocks of masses 2 kg and 1 kg move towards each other on a smooth horizontal surface with speeds 3 m/s and 6 m/s respectively. They are connected by a spring of force constant 100 N/m. Find the maximum compression of the spring.
- 0.9 m
- 0.3 m
- 0.735 m
- 0.6 m
Answer: 0.735 m
Using conservation of momentum, centre of mass velocity v_cm = (2×3 + 1×(-6))/3 = 0 m/s. Relative velocity v_rel = 9 m/s. Reduced mass μ = (2×1)/3 = 2/3 kg. Kinetic energy in CM frame = ½μ v_rel² = 27 J. This equals ½k x_max², so ½×100×x² = 27 → x = √0.54 ≈ 0.735 m.
4. A force F(x) = 4x N acts on a particle along the x-axis. What is the work done by this force as the particle moves from x = 0 to x = 3 m?
- 12 J
- 6 J
- 18 J
- 36 J
Answer: 18 J
Work done by a variable force is given by W = ∫F(x) dx from initial to final position. Here F(x) = 4x, so W = ∫₀³ 4x dx = [2x²]₀³ = 2(9) - 0 = 18 J. This is the area under the F-x graph, which is a triangle of base 3 and height 12, area = ½ × 3 × 12 = 18 J.
5. A 2 kg block slides down a rough incline of length 5 m and angle 30°. If μ = 0.2, what is its speed at the bottom? (g = 10 m/s²)
- √(100(1 - 0.2√3)) m/s
- √(50(1 - 0.2√3/2)) m/s
- √(50(1 - 0.2√3)) m/s
- √(50(1 - 0.4√3)) m/s
Answer: √(50(1 - 0.2√3)) m/s
Work-energy theorem: net work = ΔKE. Work by gravity = mgL sinθ = 2×10×5×0.5 = 50 J. Work by friction = -μ mg cosθ L = -0.2×2×10×cos30°×5 = -20×0.866 = -17.32 J. Net work = 32.68 J. So (1/2)mv² = 32.68, v² = 65.36, v = √65.36 ≈ √(50(1 - 0.2√3)) m/s.
6. For a particle moving in potential U(x) = x^3 - 3x, which point is a stable equilibrium?
- x = -1
- x = 2
- x = 0
- x = 1
Answer: x = 1
Stable equilibrium requires minimum potential energy. Set dU/dx = 3x^2 - 3 = 0 → x = ±1. Check second derivative: d²U/dx² = 6x. At x = 1, d²U/dx² = 6 > 0 (minimum, stable). At x = -1, d²U/dx² = -6 < 0 (maximum, unstable). Hence x = 1 is stable.
7. A block slides from rest down a smooth curved track that drops vertically by 5 m. What is its speed at the bottom? (g = 10 m/s²)
- 5 m/s
- 10 m/s
- 14 m/s
- 7 m/s
Answer: 10 m/s
Mechanical energy is conserved because only gravity does work (normal force does no work). Initial PE = mgh = m×10×5 = 50m J. Final KE = ½mv². Equating: ½mv² = 50m → v² = 100 → v = 10 m/s. The speed depends only on height drop, not path shape.
8. A 0.5 kg block moving at 4 m/s on a smooth horizontal surface hits a spring of stiffness 200 N/m. What is the maximum compression of the spring?
- 0.2 m
- 0.4 m
- 0.1 m
- 0.3 m
Answer: 0.2 m
By conservation of mechanical energy, initial kinetic energy converts to spring potential energy at maximum compression. KE = 0.5 * 0.5 * 4^2 = 4 J. Spring PE = 0.5 * 200 * x^2 = 100 x^2. Equating: 100 x^2 = 4 => x^2 = 0.04 => x = 0.2 m.
9. A 2 kg block slides from rest down a rough incline of length 5 m at 30° to the horizontal. The coefficient of kinetic friction is 0.2. At the bottom, it compresses a spring of force constant 100 N/m. What is the maximum compression of the spring? (g = 10 m/s²)
- 0.91 m
- 0.81 m
- 0.71 m
- 0.61 m
Answer: 0.81 m
Using work-energy theorem: initial gravitational PE = mgh = 2×10×5×sin30° = 50 J. Work done by friction = -μmg cosθ L = -0.2×2×10×cos30°×5 = -17.32 J. Net work = 50 - 17.32 = 32.68 J = (1/2)kx². So x = √(2×32.68/100) = √0.6536 ≈ 0.81 m.
10. A 2 kg block slides from rest down a rough incline of length 5 m at 30° with μ = 0.2. At the bottom it compresses a spring of k = 100 N/m. What is the maximum compression in cm? (g = 10 m/s²)
- 0.808
- 8.08
- 80.8
- 808
Answer: 80.8
Using energy conservation: mgh = ½kx² + μmg cosθ L. h = 5 sin30° = 2.5 m. mgh = 2×10×2.5 = 50 J. Friction work = μmg cosθ L = 0.2×2×10×cos30°×5 = 17.32 J. So ½kx² = 50 - 17.32 = 32.68 J, giving x = √(2×32.68/100) = 0.808 m = 80.8 cm.
11. A block slides from rest down a rough incline of length 5 m at 30° to the horizontal, then enters a smooth vertical loop of radius 2 m. The coefficient of friction on the incline is 0.2. What minimum height H (measured from the bottom of the loop) must the block start from to just complete the loop? (Use g = 10 m/s², √10 ≈ 3.16, √20 ≈ 4.47)
- 6.73 m
- 5.00 m
- 5.87 m
- 4.13 m
Answer: 5.87 m
Using energy conservation with friction: initial PE = mgH, friction work = -μmg cosθ L, final energy at loop top = mg(2R) + ½m(gR). Substituting values: 10H - 0.2×10×cos30°×5 = 10×4 + 5×2 → 10H - 8.66 = 50 → H = 5.866 m ≈ 5.87 m.
12. A particle moves in a potential U(x) = x^2 - 4x (J). Its total mechanical energy is 5 J. What are the turning points?
- x = -1 m and x = 3 m
- x = 1 m and x = 3 m
- x = -1 m and x = 5 m
- x = 1 m and x = 5 m
Answer: x = -1 m and x = 5 m
Turning points occur where kinetic energy is zero, so potential energy equals total energy: U(x) = E. Solve x^2 - 4x = 5 => x^2 - 4x - 5 = 0 => (x-5)(x+1)=0 => x = -1 m and x = 5 m. The particle oscillates between these points.