A force of 4.9 N acts on a 14 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.

Explanation:

The work done on the body = force applied x displacement

in this case , the acceleration of body a = = = 0.35 ms⁻²

The displacement in first second S₁ = u + a x t²

here u = 0 , because body was at rest

Thus S₁ = x 0.35 x ( 1 )² = = 0.175 m

The work done = 4.9 x 0.175 =  0.86 J

The displacement in 2 seconds = x 0.35 x ( 2 )² = 0.7 m

Work done in 2 seconds = 4.9 x 0.7 = 3.43 J

Work done in second second = 3.43 - 0.86 = 2.57 J

The displacement in three seconds = x 0.35 x ( 3 )² = 1.575 m

Work done in three seconds = 4.9 x 1.575 = 7.7 J

Work done in third second = 7.7 - 3.43 = 4.3 J

The power at the end of third second is 4.3 watt

Because 4.3 J of work is done in the last second .

a and b sometimes have a positive value

double the distance = divide strength by 4