Question

A block of mass 10 kg rests on a horizontal floor. The acceleraton due to gravity is $9.81\frac{m}{s^2}$ . The coefficient of static friction between the floor and the block is 0.2. A horizontal forcce of 10 N is applied on the block as shown in the figure. The magnitude of force of friction (in N) on the block is

Answer 1

$W=mg=10\times 9.81=98.1N$
Normal force, N $N=W=98.1N$
Now, maximum value of static friction force
$={\mu }_{s}N=0.2\times 98.1=19.62N$
$\because$ Applied force is less than static friction force,so friction force will be same as applied force. Hence, Friction force = 10 N