Numerical Problems: Work, Energy and Power

Q1. Force acting on a particle moving in a straight line varies with the velocity of the particle as F = K / v. Here K is constant. The work done by this force in time t is
Sol: 
Given F = K / v.
By Newton's second law, m dv/dt = F = K / v.
Rearrange: v dv = (K / m) dt.
Integrate both sides from t = 0 (v = 0) to t (v = v): ∫ v dv = (K / m) ∫ dt.
Thus v²/2 = (K / m) t.
Multiply both sides by m/2: (1/2) m v² = K t.
Since the work done by the force in time t equals the change in kinetic energy, W = K t.
Hence the work done in time t is K t.

Q2. A particle of mass 0.5 kg is displaced from position r₁ = (2î + 3ĵ + 1k̂) to r₂ = (4î + 3ĵ + 2k̂) by applying a force of magnitude 30 N acting along the direction (î + ĵ + k̂). Find the work done by the force.
Sol: 
Magnitude of force, F = 30 N.
Unit vector in the given direction: f̂ = (î + ĵ + k̂) / √3.
Force vector: F = 30 f̂ = 30 (î + ĵ + k̂) / √3 = 10√3 (î + ĵ + k̂).
Displacement S = r₂ - r₁ = (4-2)î + (3-3)ĵ + (2-1)k̂ = 2î + 0ĵ + 1k̂.
Work done W = F · S = 10√3 (î + ĵ + k̂) · (2î + k̂) = 10√3 (2 + 0 + 1) = 30√3 J.
Therefore the work done by the force is 30√3 J.

Q3: The potential energy of a particle of mass 5 kg moving in the x-y plane is given by U = (-7x + 24y) J where x and y being in metre. If the particle starts from rest from origin, then speed of particle at t = 2s is
Sol: 
Force is given by the negative gradient of U:
F = -∇U = -[(∂U/∂x) î + (∂U/∂y) ĵ].
Compute partial derivatives: ∂U/∂x = -7, ∂U/∂y = 24.
Thus F = -[(-7) î + 24 ĵ] = 7 î - 24 ĵ.
Magnitude of the resultant force: |F| = √(7² + (-24)²) = √(49 + 576) = 25 N.
Acceleration a = F/m = 25 / 5 = 5 m/s² (constant).
Starting from rest (u = 0), speed after t = 2 s: v = u + at = 0 + 5 × 2 = 10 m/s.
Hence the speed at t = 2 s is 10 m/s.

Q4: If the engine power is 3.3kW and it is 60% efficient, how much water will it pump in 5s from a height of 10m?
Sol: 
Engine power P_engine = 3.3 kW = 3300 W.
Useful (pump) power = efficiency × P_engine = 0.60 × 3300 = 1980 W.
Energy required to lift mass m by height h in time t: P_useful = (m g h) / t ⇒ m = (P_useful × t) / (g h).
Take g = 10 m/s² and h = 10 m, t = 5 s.
m = (1980 × 5) / (10 × 10) = 9900 / 100 = 99 kg.
Therefore the engine will pump 99 kg of water in 5 s (about 99 litres).

Q5: A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is
Sol: 
Initial height = 100 m; final height = 20 m. Loss in gravitational potential energy = mg(100 - 20) = mg × 80.
By conservation of mechanical energy (no friction), this loss equals the kinetic energy at the final level:
(1/2) m v² = m g × 80 ⇒ v² = 2 g × 80.
With g = 10 m/s²: v² = 1600 ⇒ v = 40 m/s.
Hence the speed of the ball at the horizontal base (height 20 m) is 40 m/s.

Q6: A bullet of mass 0.01 kg, travelling at a speed of 500 m/s^-1, strikes a block of mass 2 kg, which is suspended by a string of length 5 m, and emerges out. The block rises by a vertical distance of 0.1 m. The speed of the bullet after it emerges from the block is
Sol:
Mass of bullet m = 0.01 kg; initial bullet speed u = 500 m/s.
Mass of block M = 2 kg. Block rises by h = 0.1 m after collision.
Velocity of block immediately after collision, V_block, from energy: M g h = (1/2) M V_block² ⇒ V_block = √(2 g h).
Using g = 9.8 m/s²: V_block = √(2 × 9.8 × 0.1) = √1.96 = 1.4 m/s.
Apply conservation of linear momentum for the collision (bullet emerges):
m u = m v' + M V_block, where v' is bullet speed after emerging.
Substitute values: 0.01 × 500 = 0.01 v' + 2 × 1.4 ⇒ 5 = 0.01 v' + 2.8.
Hence 0.01 v' = 2.2 ⇒ v' = 2.2 / 0.01 = 220 m/s.
Therefore the speed of the bullet after emerging is 220 m/s.

Q7: One end of an unstretched vertical spring is attached to the ceiling and an object attached to the other end is slowly lowered to its equilibrium position. If S be gain in spring energy & G be loss in gravitational potential energy in the process, then
Sol: 
At equilibrium the extension x satisfies k x = m g ⇒ x = m g / k.
Loss in gravitational potential energy when lowered by x: G = m g x.
Energy stored in the spring at equilibrium: S = (1/2) k x².
But k x = m g, so S = (1/2) (k x) x = (1/2) (m g) x = (1/2) G.
Therefore G = 2 S.

Q8: A vertical spring of force constant 100 N/m is attached with a hanging mass of 10 kg. Now an external force is applied on the mass so that the spring is stretched by additional 2 m. The work done by the force F is : (g = 10 m/s²)

Sol:
Spring constant k = 100 N/m, mass m = 10 kg, g = 10 m/s².
Equilibrium extension x_i from natural length: x_i = m g / k = (10 × 10) / 100 = 1 m.
Additional extension = 2 m ⇒ final extension x_f = 1 + 2 = 3 m.
Work done by external force (quasi-static) equals increase in spring potential plus change in gravitational potential:
W_ext = ΔU_spring + ΔU_gravity.
ΔU_spring = (1/2) k (x_f² - x_i²) = (1/2) × 100 × (9 - 1) = 50 × 8 = 400 J.
ΔU_gravity = m g (vertical rise of mass). The mass moves downward by 2 m, so its gravitational potential decreases by m g × 2 = 10 × 10 × 2 = 200 J ⇒ ΔU_gravity = -200 J.
Thus W_ext = 400 + (-200) = 200 J.
Therefore the work done by the external force is 200 J.

Q9: A block of mass 5 kg is released from rest at a height of 20 m on a smooth inclined plane making an angle of 30° with the horizontal. The block slides down and reaches the bottom. If the coefficient of kinetic friction between the block and a rough horizontal surface at the bottom is 0.2, and the horizontal distance traveled is 10 m, find the total work done by all forces. (g = 10 m/s²)
Sol:
Initial potential energy at height 20 m: U = m g h = 5 × 10 × 20 = 1000 J.
Since the incline is smooth, all this potential energy is converted to kinetic energy at the bottom of the incline: K_bottom = 1000 J.
Work done by kinetic friction on the horizontal surface over distance d = 10 m: W_f = -μ_k m g d = -0.2 × 5 × 10 × 10 = -100 J (negative because friction removes energy).
After travelling the horizontal distance, final kinetic energy K_final = K_bottom + W_f = 1000 - 100 = 900 J.
Total work done by all forces for the whole motion (from start to after 10 m on horizontal) equals change in kinetic energy: ΔK = K_final - K_initial = 900 - 0 = 900 J.
Therefore the total work done by all forces is 900 J.

Q10. A 2 kg particle is moving in a straight line under a force F = (6x - 4) N, where x is the displacement in meters. If the particle starts from rest at x = 0, calculate the kinetic energy of the particle when it has moved 5 m. (g = 10 m/s²)
Sol: 
Work done by the force from x = 0 to x = 5 m:
W = ∫₀⁵ F dx = ∫₀⁵ (6x - 4) dx.
Integrate: W = [3 x² - 4 x]₀⁵ = (3 × 25 - 4 × 5) - 0 = 75 - 20 = 55 J.
By the work-energy theorem, the work done equals the change in kinetic energy. Initial KE = 0, so KE at x = 5 m = 55 J.
Therefore the kinetic energy after 5 m is 55 J.