Question

In the figure shown, what is the value of mass ‘$m$’ such that block $A$ slides up with a constant velocity. Take $g=10{\text{m/s}}^2$, Coefficient of friction between inclined plane and block $A$ is $0.5$.

• A. $\text{1 kg}$
• B. $\text{2 kg}$
• C. $\text{3 kg}$
• D. $\text{4 kg}$

Answer 1

The correct option is A $\text{1 kg}$



On the $\text{1 kg}$ block, the frictional force will act down the incline plane opposite to the relative motion of the block on the surface.
For blocks to move with constant velocity, net force has to be zero.

Using FBD

For $m$ :
$T=mg$, the maximum tension on the rope

For $\text{1 kg}$:
$f=\mu N=\mu \left(1\right)g\cos \theta =\mu g\cos \theta$
$T=\left(1\right)g\sin \theta +f=g\sin \theta +\frac{1}{2}g\cos \theta$
$=g[\frac{3}{5}+\left(\frac{1}{2}\right)\frac{4}{5}]$
$=g=10N$
(As $\theta ={37}^{\circ }$, $\sin \theta =\frac{3}{5},\cos \theta =\frac{4}{5}$)
Hence, $T=mg$ = $\text{10 N}$ $\Rightarrow$ $m=\text{1 kg}$.