JEE Main Physics question
A bullet of mass 0.05 kg moving with a speed of 80 ms⁻¹ enters a wooden block and is stopped after a distance of 0.40 m. The average resistive force exerted by the block on the bullet is
Answer: (C)
Solution
By the work-energy theorem, F·s = initial KE ⟹ F = (½ × 0.05 × 80²)/0.40 = 160/0.40 = 400 N.