Work, Energy and Power
ICSE · Class 11 · Physics
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A force F = (3x² + 2x) N acts on a particle along the x-axis. What is the work done by the force as the particle moves from x = 1 m to x = 3 m?
Two bodies A and B of masses m and 3m respectively have the same kinetic energy. They are brought to rest by applying the same retarding force. The ratio of distances covered by A to B before coming to rest is:
A particle moves along the x-axis under a potential energy U(x) = (x² – 4x) J, where x is in metres. At which position is the particle in stable equilibrium and what is the minimum potential energy?
A ball of mass 0.5 kg is dropped from a height of 5 m on a floor. The coefficient of restitution is 0.8. What is the height attained by the ball after the second bounce? (Take g = 10 m/s²)
Sample Questions
In a head-on elastic collision, a body of mass m₁ = 1 kg moving at 6 m/s collides with a stationary body of mass m₂ = 3 kg. What fraction of the initial kinetic energy is transferred to the second body?
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3/4
Step 1: Use the formula for fraction of KE transferred in elastic collision (u₂ = 0): KE_transferred/KE_initial = 4m₁m₂/(m₁+m₂)². Step 2: Here m₁ = 1 kg, m₂ = 3 kg. Substituting: = 4(1)(3)/(1+3)² = 12/16 = 3/4. Step 3: Verify using velocities: v₂ = 2m₁u₁/(m₁+m₂) = 2(1)(6)/4 = 3 m/s. KE transferred = (1/2)(3)(3²) = 13.5 J. Initial KE = (1/2)(1)(6²) = 18 J. Fraction = 13.5/18 = 3/4 ✓. Step 4: Also, v₁ = (m₁–m₂)/(m₁+m₂)×u₁ = (1–3)/4 × 6 = –3 m/s (rebounds). KE retained = (1/2)(1)(9) = 4.5 J = 1/4 of initial KE. Step 5: Maximum KE transfer occurs when m₁ = m₂ (fraction = 1, i.e., 100% transfer).
The potential energy of a spring-mass system at maximum displacement x₀ is 8 J. When the displacement is x₀/2, the kinetic energy of the system is:
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6 J
Step 1: At maximum displacement x₀, velocity = 0, so all energy is potential: PE_max = (1/2)kx₀² = 8 J → Total mechanical energy E = 8 J. Step 2: At displacement x = x₀/2, PE = (1/2)k(x₀/2)² = (1/2)k(x₀²/4) = (1/4)×(1/2)kx₀² = (1/4) × 8 = 2 J. Step 3: By conservation of mechanical energy: KE + PE = E (total) → KE = E – PE = 8 – 2 = 6 J. Step 4: Common error: students often think KE = E/2 = 4 J (confusing with mean position where x = 0). At mean position KE = 8 J (maximum). Step 5: Note that PE is proportional to x², so halving the displacement reduces PE to 1/4th of its maximum, not 1/2.
An engine of power 9 kW pulls a train of mass 10,000 kg on a level track at a uniform speed. If the coefficient of friction is 0.02, what is the speed of the train? (Take g = 10 m/s²)
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4.5 m/s
Step 1: At uniform speed, the driving force equals the frictional (retarding) force. Step 2: Frictional force F = μ × mg = 0.02 × 10,000 × 10 = 2000 N. Step 3: Power P = F × v → v = P/F. Step 4: v = 9000/2000 = 4.5 m/s. Step 5: Common error: using P = mv²/2 (which gives KE, not power at constant speed). At uniform speed, power = Force × velocity, and the only force to overcome is friction on a level track. If the track were inclined, we would also include the component of gravity along the incline.
A bullet of mass 20 g moving at 500 m/s strikes a wooden block of mass 980 g at rest and gets embedded in it. What percentage of the initial kinetic energy is lost in the collision?
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98%
Step 1: This is a perfectly inelastic collision. m₁ = 0.02 kg, u₁ = 500 m/s, m₂ = 0.98 kg, u₂ = 0. Step 2: Common velocity after collision: v = m₁u₁/(m₁+m₂) = (0.02×500)/(0.02+0.98) = 10/1 = 10 m/s. Step 3: Initial KE = (1/2)(0.02)(500²) = (1/2)(0.02)(250000) = 2500 J. Step 4: Final KE = (1/2)(1)(10²) = 50 J. Loss = 2500 – 50 = 2450 J. Percentage loss = (2450/2500) × 100 = 98%. Step 5: Alternatively, use the formula: KE_loss/KE_initial = m₂/(m₁+m₂) = 0.98/1.00 = 0.98 = 98% (valid when m₂ is initially at rest). This shows that when a very light body embeds in a very heavy body, almost all kinet
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