Question

A block of mass $10\text{kg}$ is kept on a rough inclined plane as shown in the figure. A force of $3\text{N}$ is applied on the block. The coefficient of static friction between the plane and the block is $0.6$. What should be the minimum value of force $P$, such that the block does not move downward? $\text{(take}g=10{\text{ms}}^{-2})$

• A. $32\text{N}$
• B. $18\text{N}$
• C. $23\text{N}$
• D. $25\text{N}$

Answer 1

The correct option is A $32\text{N}$


When minimum force $P$ is applied to stop the downward motion of the block,

$3+mg\sin \theta =P+\mu mg\cos \theta$

$3+\frac{100}{\sqrt{2}}=P+0.6\times 100\times \frac{1}{\sqrt{2}}$

$3+50\sqrt{2}=P+30\sqrt{2}$

$\therefore P=31.28\approx 32N$

Hence, option $\left(A\right)$ is correct.