Question

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

• A. $40\text{N}$
• B. $30\text{N}$
• C. $60\text{N}$
• D. $50\text{N}$

Answer 1

The correct option is D $50\text{N}$

$\text{FBD}$ of block-

From vertical equilibrium,
$N=mg=4\times 10=40\text{N}$
Limiting value of friction,
$F_f=\mu N=0.8\times 40=32\text{N}$
Since, the value of limiting friction is greater than external force, therefore, the block will not slide and
$F_f=F_e=30\text{N}$
Now, the resultant force exerted by ground will be,
$R=\sqrt{N^2+F_f^2}$

$\Rightarrow R=\sqrt{{40}^2+{30}^2}$
$\Rightarrow R=50\text{N}$