JEE Main Physics question
A body is coming with a velocity of 72 km h⁻¹ on a rough horizontal surface of coefficient of friction 0.5. If the acceleration due to gravity is 10 ms⁻², find the minimum distance in which it can be stopped.
Answer: (B)
Solution
v = 20 ms⁻¹, a = μg = 5 ms⁻². s = v²/2a = 400/10 = 40 m.