Question

A block of mass ${}^{\prime }{m}^{\prime }$ is placed on an inclined plane having coefficient of friction ${}^{\prime }{m}^{\prime }$. The plane is making an angle $\theta$ with the horizontal. The minimum value of upward force acting along the inclined plane that can just move the block up is

• A. $mg\cos \theta$
• B. $mmg\cos \theta$
• C. $mg\sin \theta$
• D. $mmg\sin \theta$

Answer 1

Answer: (C)

Suppose a force $F$ is acting on the block as shown in the figure.
Minimum value of the force required to just move the block up is
$F_{\min }=mg\sin \theta +f_{\min }$
Where, $f=$ frictional force $f_{\min }=0$
$\Rightarrow F_{\min }=mg\sin \theta$