Question

A block sliding down a rough ${45}^o$ inclined plane has half the velocity it would have had, the inclined plane been smooth. The coefficient of sliding friction between block and the inclined plane is

• A. $\frac{1}{4}$
• B. $\frac{3}{4}$
• C. $\frac{1}{2\sqrt{2}}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (B)

$\frac{3}{4}$

We consider the case with no friction. In this case, the acceleration $a_{smooth}=g\sin {45}^{\circ }=\frac{g}{\sqrt{2}}$

Using 2nd equation of motion, $(v_{smooth}=\sqrt{2\times a_{smooth}\times l})$ where $l$ is the length of inclined plane

We are given that $v_{rough}=\sqrt{2\times a_{rough}\times l}=\frac{1}{2}{v}_{smooth}=\frac{1}{2}\sqrt{2\times a_{smooth}\times l}$.

Clearly, this means $a_{rough}=\frac{1}{4}{a}_{smooth}$.

Now, consider the case of the rough inclined plane. Net force down the incline is given by:

$mg\sin {45}^{\circ }-f=m{a}_{rough}$, where $f$ is the limiting friction.

Thus, $\frac{mg}{\sqrt{2}}-\mu \frac{mg}{\sqrt{2}}=\frac{m{a}_{smooth}}{4}=\frac{mg}{4\sqrt{2}}$

Upon solving the above, we get $\mu =3/4$

$\to QED.$