A block of mass m=2 ~kg is initially at rest on a horizontal surface. A horizontal force F ₁=(6 ~N ) i and a vertical force F ₂=(10 ~N ) j are then applied to the block. The coefficients of static friction and kinetic friction for the block and the surfaces are 0.4 and 0.25 , respectively. The magnitude of the frictional force acting on the block is (assume, g=10 ~m / s ^2 )

Step-by-step solution

In addition to the forces shown in figure, there will be another upcoming normal force F _N exerted by the floor on the block. Using Newton's second law in horizontal and vertical directions, we get In horizontal direction, F₁-f_s=m a ( i ) and in vertical direction, F₂+F_N-m g=0 ( ii ) Given, F₂=10 ~N , m=2 ~kg and g=10 ~ms ⁻² 10+F_N-m g=0 or F_N=(2 10)-10=10 ~N ( iii ) Hence, static friction f_s = _s F_N=0.4 10=4 ~N ( iv ) Clearly, it is lesser than static force F₁(=6 ~N ) , so block will move, i.e. we are dealin