Work of 2 Joules is done in stretching a spring from its natural length to 14 c m beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (14 c m )

F =28.57 N

Explanation:

Assuming that the spring is ideally followed  Hooke's law, This means

F=kx

Here F is the force required to stretch or compress the spring through x distance.

the amount of the work done in the displacement of the spring would be:

W = ∫F.dx

=∫kx dx

=1/2kx^2

=1/2k×0.14^2

⇒ 0.0098k =2

⇒ k = 204.081

Therefore, the force (in Newtons) that holds the spring stretched at the same distance.

F = 204.081×0.14

F =28.57 N

power out to power in less than 100%

this would be about 43%