STANDARD Physics Work done by a constant force MCQ Question
A block of mass 2 kg initially at rest moves under the action of an applied horizontal force of 6 N on a rough horizontal surface. The coefficient of friction between block and surface is 0.1. The work done by the applied force in 10 s is (Take g = 10 m s⁻²)
200 J
−200 J
600 J
−600 J
Correct Answer
Detailed Explanation
The force of friction is calculated as f = μN = 0.1 × 2 kg × 10 m s⁻² = 2 N. The net force is F' = F - f = 6 N - 2 N = 4 N. The acceleration is a = F'/m = 4 N / 2 kg = 2 m s⁻². The distance traveled in 10 s is d = 0.5 × a × t² = 0.5 × 2 m s⁻² × (10 s)² = 100 m. The work done is W = F × d × cos θ = 6 N × 100 m × cos 0° = 600 J.
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