Question

A block of mass $2 \, kg$ slides on an inclined plane which makes an angle of $30^\circ $ with the horizontal. The coefficient of friction between the block and the surface is $\sqrt{\frac{3}{2}}$ . What force along the plane should be applied to the block so that it moves down?

• A. $11.21$
• B. $4.5$
• C. $6.7$
• D. $30.14$

Answer 1

Answer: (A)

$mgsin\theta =\left(\right.2\left.\right)\left(\right.10\left.\right)$ $\left(\right. \frac{1}{2} \left.\right)$ $=10N=F_{1}$ (say)

$\mu mg\,cos\theta =$ $\left(\right. \sqrt{\frac{3}{2}} \left.\right)$ $\left(\right.2\left.\right)\left(\right.10\left.\right)$ $\left(\right. \frac{\sqrt{3}}{2} \left.\right)$ $=21.21N=F_{2}$ (say)
Force required to move the block down the plane with constant velocity.

$F _{1}$ will be acting downwards, while $F _{2}$ upwards.
Since $F_{2}>F_{1}$, force required
$F=F_{2}-F_{1}=11.21 N$