Question

If a body have kinetic energy T, moving on a rough horizontal surface stops at distance y. The frictional force exerted on the body is

(a) T/√Y

(b) √T/Y

(c) YT

(d) T/Y

Answer 1

Correct option  (d) T/Y

Explanation:

Let the initial velocity be u and the coefficient of friction be μ.

Now frictional force is responsible for retardation of the body

So, retardation due to friction (a) = μg Using laws of kinematics

where y is the distance body moved before stopping

where f is the force of friction