Question

A block of mass m =2kg is placed in equilibrium on moving planck acceleration with acceleration a=4m/s if coefficient of friction between, is 0.2 the friction force acting on block id

Answer 1

Answer:

3.924 N

Explanation:

Given data in the question :

Mass of the block, m = 2 kg

acceleration of the plank = 4 m/s²

coefficient of friction between, μ = 0.2

It is also given that the block is in equilibrium with the plank.

Now,

the frictional force is given as:

f = μN

where, N is the normal reaction

N = mg

g is the acceleration due to the gravity

thus,

N = 2 × 9.81 = 19.62 Newton

hence,

the friction force acting on the block is

f = 0.2 × 19.62 = 3.924 N