Question

a block of mass 2 kg initially at rest moves under the action of an applied horizontal force of 6 Newton on a rough horizontal surface the coefficient of friction between block and services. 1 the work done by the force in 10 second is ​

Answer 1

Explanation:

M = 2 kg

Applied force F = 7 N

Coefficient of kinetic friction μ = 0.1

Normal reaction is N = mg = 2 x 9.8 = 19.6 N

Hence, force or friction is f = μN = 1.96 N

Total force = F - f = 7 - 1.96 = 5.04 N

Acceleration of body is

   

Displacement of body in time t is

In t = 10 s

(a) Work done by F is

     W = F x = 7 x 125 J = 875 J

(b) Work done by friction = - f x

                                   = - 1.96 x 125 = -245 J

(c) Net work done = (875 - 245) = 630 J

(d) Change in kinetic energy = Net work done = 630 J