Question

A block 'A' of mass $4kg$ is kept on ground. The coefficient of friction between the block and the ground is $0.8$. The external force of magnitude $30N$ is applied parallel to the ground. The resultant force exerted by the ground on the block is: $(g=10m{s}^{-2})$

• A. $40N$
• B. $30N$
• C. $zero$
• D. $50N$

Answer 1

The correct option is D $50N$

Normal force on the block, $N=mg=4\times 10=40N$
Maximum friction force which can act on block $=\mu N=0.8\times 40=32N$
As the force applied is less than the maximum friction force so friction force on block is 30N (to stop relative motion)
Both friction force and normal force are exerted by ground and their direction is horizontal and vertical respectively (i.e perpendicular to each other). Hence, net force exerted by ground is $\sqrt{{30}^2+{40}^2}=50N$