A block of mass m=2 Kg is initially at rest oil a horizontal surface. A horizontal force vecF1=(6 N ) hati and a vertical force vec F 2=(10 N ) hatj 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 )

Question

A block of mass m=2 Kg is initially at rest oil a horizontal surface. A horizontal force vecF1=(6 N ) hati and a vertical force vec F 2=(10 N ) hatj 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

Tardigrade Admin · Accepted Answer

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} = ma \dots (i)

and in vertical direction,

F_{2} + F_{N} - m g = 0 \dots (ii)

Given,

F_{2} = 10 N, m = 2 k g

and

g = 10 m s^{- 2}

\Rightarrow 10 + F_{N} - m g = 0

or

F_{N} = (2 \times 10) - 10 = 10 N \dots

(iii)
Hence, static friction,

f_{s} = μ_{s} F_{N}

= 0.4 \times 10 = 4 N \dots (i v)

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} = μ_{k} F_{N} = 0.25 \times 10 = 2.5 N

$2.5 N$

$4.0 N$

$3.3 N$

$3.0 N$

2.5N

4.0N

3.3N

3.0N

$2.5 N$

$4.0 N$

$3.3 N$

$3.0 N$