Question

A block of mass $m=2\, Kg$ is initially at rest oil a horizontal surface. A horizontal force $\vec{F}_{1}=(6\,N ) \hat{i}$ and a vertical force $\vec{ F _{2}}=(10 \,N ) \hat{j}$ are then applied to the block. The coefficients of static friction and kinetic friction for the block and the surface are $0.4$ and $0.25$ respectively. The magnitude of the frictional force acting on the block is
(Assume $g =10\, m / s ^{2}$ )

• A. $2.5\, N$
• B. $4.0 \,N$
• C. $3.3 \,N$
• D. $3.0 \,N$

Answer 1

Answer: (A)

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,
$F_{1}-f_{s}=m a \ldots (i) $
and in vertical direction,
$F_{2}+F_{N}-m g=0 \ldots (ii) $
Given, $F_{2}=10 \,N , m=2\, kg$ and $g=10\, ms ^{-2}$
$\Rightarrow 10+F_{N}-m g=0$
or $F_{N}=(2 \times 10)-10=10 N \ldots$ (iii)
Hence, static friction, $f_{s}=\mu_{s} F_{N}$
$=0.4 \times 10=4 N \ldots (iv) $
Clearly, it is lesser than static force $F_{1}(=6 N)$,
so block will move, i.e. we are dealing with kinetic friction now.
Hence, friction force (kinetic) acting on the body is
$f_{k}=\mu_{k} F_{N}=0.25 \times 10=2.5 \,N$