Question

A body having kinetic energy $k$ moving on a rough horizontal surface is stopped at a distance $x$ by constant frictional force. The force of friction exerted on the body is

• A. $\frac{k}{x}$
• B. $\frac{\sqrt{k}}{x}$
• C. $\frac{k}{\sqrt{x}}$
• D. $kx$

Answer 1

The correct option is A $\frac{k}{x}$

Let, m be the mass of the body, v be the initial speed of the body, F be the force of friction and a be the retardation.
From work energy theorm
$\Rightarrow Fx=\frac{1}{2}m{v}^2$
Here, kinetic energy of a body is k, and $k=\frac{1}{2}m{v}^2$
Therefore,
$Fx=k$
$F=\frac{k}{x}$