Question

A block of mass $10\text{kg}$ is slowly moved up on a smooth inclined plane of inclination $\theta$ by a person. Work done by the person to move the block through a distance of $2m$ is $120J$. If force is applied in horizontal direction. Find out inclination angle $\theta$. (Take $g=10m/s^2$)

• A. ${53}^{\circ }$
• B. ${45}^{\circ }$
• C. ${37}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is C ${37}^{\circ }$


Displacement, $S=2\text{m}$
When force is applied horizontally, as block moves slowly,
$F\cos \theta =mg\sin \theta$
$F=mg\frac{\sin \theta }{\cos \theta }$
Work done by force,
$W=F.S$
$W=FS\cos \theta$
$=\frac{mg\sin \theta }{\cos \theta }2\times \cos \theta$
$120=10\times 10\times 2\times \sin \theta$
$120=200\sin \theta$
$\sin \theta =\frac{3}{5}$
$\Rightarrow \theta ={37}^{\circ }$