JEE Main Physics question
A force of 4 N is applied on a body of mass 20 kg, initially at rest. The work done by this force in the 3rd second is
Answer: (A)
Solution
a = F/m = 4/20 = 0.2 ms⁻². Distance in the 3rd second, s₃ = (a/2)(2×3−1) = 0.1×5 = 0.5 m. Work in that second = F·s₃ = 4 × 0.5 = 2 J.