Question

A body of mass m, having momentum $p$ is moving on a rough horizontal surface. If it is stopped in a distance $x$, the coefficient of friction between the body and the surface is:

• A. $\mu =\frac{\rho }{2mgx}$
• B. $\mu =\frac{{\rho }^2}{2mgx}$
• C. $\mu =\frac{{\rho }^2}{2g{m}^{2}x}$
• D. $\mu =\frac{{\rho }^2}{2g{m}^2{x}^2}$

Answer 1

Answer: (C)

$\mu =\frac{{\rho }^2}{2g{m}^{2}x}$

Frictional force $=\mu mg$

So, Retardation $=a=\frac{\mu mg}{m}=\mu g$

Now, $v^2=2ax=2\mu gx$

or, $m^2{v}^2=2m^2\mu gx$

or, $p^2=2m^2\mu gh$

or, $\mu =\frac{p^2}{2g{m}^{2}x}$