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. $\frac{T}{\sqrt{y}}$
• B. $\frac{\sqrt{T}}{y}$
• C. $yT$
• D. $\frac{T}{y}$

Answer 1

The correct option is D $\frac{T}{y}$

A body of mass m moving initially with velocity u on a rough surface and due to friction, it comes to rest after covering a distance s, then the retarding force is given by

$ma=\mu mg$ or $a=\mu g$ ...(i)

Using, $v^2=u^2-2as$, we have

$0=u^2-2\left(\mu g\right)s$ ...(ii)

$T=\frac{1}{2}m{u}^2$ or $u^2=\frac{2T}{m}$

We have from Eq. (ii),

$0=\frac{2T}{m}-2\left(\mu g\right)s$

$\frac{2T}{m}=2\mu gy$ $[\because s=y]$ ...(iii)

Frictional force, $f=\mu mg$ ...(iv)

From Eqs. (iii) and (iv), we get $f=\frac{T}{y}$