A block of mass $5 kg$ is at rest on a rough inclined surface. If angle of inclination of plane is $60^{\circ}$, then force applied by it on block is .......... $N$
A$50$
B$25 \sqrt{3}$
C$25$
D$50 \sqrt{3}$
Medium
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A$50$
a (a)
We need to find the resultant force between $m g \sin \theta$ and $m g \cos \theta$ which is actually $m g$ then $m g=(5 \times 10)\,N =50\,N$.
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