Numerical Problems: Work, Energy and Power | Physics Class 11 - NEET PDF Download

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: 
⇒ F = K / v
⇒ m dv / dt = K / v
⇒ v dv = K / m dt
⇒ ∫ v dv = (K / m) ∫ dt
⇒ v² / 2 = (K / m) t
⇒ 1 / 2 m v² = K t ⇒ W = 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 force direction, f̂ = (î + ĵ + k̂)/√3
Force vector: F = F·f̂ = 30 × (î + ĵ + k̂)/√3 = 10√3 (î + ĵ + k̂)
S = r₂ - r₁ = (4-2)î + (3-3)ĵ + (2-1)k̂ = 2î + k̂
W = F · S = [10√3 (î + ĵ + k̂)] · [2î + k̂]
W = 10√3 × (2 + 0 + 1) = 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: The force F is given by the negative gradient of the potential energy U:
F = - [(∂U/∂x) i + (∂U/∂y) j] = - [(-7) i + (24) j]
F = √(7² + 24²) = √(49 + 576) = 25 N
Acceleration a = F/m = 25 / 5 = 5 m/s²
Velocity v = u + at, where u = 0 (starts from rest)
v = 0 + 5 × 2 = 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: Power used to pump the water,
mgh / t = 100 × 10 × 5 = 2000W
Thus, Power of engine = 2000 × 100 / 60 = 3.3kW

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: Loss in potential energy = Gain in KE
mg × 80 = ¹/₂ mv²
⇒ v² = 2 × 10 × 80 = 1600
⇒ v = 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: By law of Conservation of Linear momentum
mu = mV + MV ...(1)
where m = mass of bullet
M = mass of block
u = velocity of bullet before collision
V = velocity of bullet after collision
V = velocity of block after collision
By Conservation of Energy
Mgh = ¹/₂ MV²
⇒ V = √(2 × 9.8 × 0.1)
⇒ V = 1.4 m/s
Put in (1), we get
5 = 0.01V + 2(1.4) ⇒ V = ²/_0.01
⇒ V = 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 position x = ^mg/_k
U_spring = ¹/₂ kx² = ¹/₂ k (^mg/_k)x = ^mgx/₂ = ¹/₂ (loss in GPE) ⇒ G = 2S

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: At equilibrium, mg = kx₀ ⇒ x₀ = ^mg/_k = ^{10 × 10}/₁₀₀ = 1 m
W_ext = U_f - U_i
= ¹/₂ k(x²_f) - [¹/₂ k(x²_i) + mgh]
= ¹/₂ k(x²_f - x²_i) - mgh
= ¹/₂ × 100 × (3² - 1²) - 10 × 10 × 2
= 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 = mgh = 5 × 10 × 20 = 1000 J
Work done against friction on the horizontal surface = μmg × d = 0.2 × 5 × 10 × 10 = 100 J
Using work-energy theorem, total work done = change in kinetic energy + work against friction
Kinetic energy at bottom of incline = 1000 J (conserved as incline is smooth)
Total work done = 1000 J - 100 J = 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 = ∫F dx from 0 to 5
F = 6x - 4
Work = ∫(6x - 4) dx = [3x² - 4x] from 0 to 5 = (3×25 - 4×5) - 0 = 75 - 20 = 55 J
By work-energy theorem, work done = change in kinetic energy
Initial KE = 0, so kinetic energy at 5 m = 55 J