Question

In the arrangement shown in Fig. there is no friction between the block of mass $2m$ and ground, but there is friction between of the blocks of masses m and $2m$. The block of mass m is stationary with respect to block of mass $2m$. The value of coefficient of friction between m and $2m$ is:

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

Answer 1

Answer: (C)

$\frac{1}{4}$

REF.Image.

Net acceleration = $\frac{DrivingForce}{TotalMass}$

$a=\frac{mg}{4m}\Rightarrow a=g/4$

As mass m B stationary w.r.t 2 m, friction

should provide this effect.

$mmg=mg/4$

$m=1/4$