Question

A block of mass 4kg is kept on a horizontal surface with coefficient of friction between the surface as 0.2 Find the magnitude of frictional force acting:

Answer 1

Answer:

force of friction = u R

= u mg

= 0.2*4*10

=8 N

Answer 2

Answer:

Magnitude of the frictional force acting is, f = 7.84 N

Explanation:

In the question,

Mass of the block, m = 4 kg

Co-efficient of the friction between the surfaces, μ = 0.2

Now,

We know that the Magnitude of the frictional force is given by,

$f=\mu N$

where,

N is the Normal force on the surface.

Now,

$N=mgN=(4)(9.8)\\N=39.2$

So,

$f=\mu N\\f=(0.2)\times 39.2\\f=7.84 N$

Therefore, the magnitude of the frictional force acting is, f = 7.84 N