A certain coiled spring with an unstretched length of 20cm required a force of 2N to stretch it by 0.2cm. What work is done in stretching it by 2cm if the elastic limit is not exceeded?

Answer 1

k = 2 N / .002 m = 1 kN / m

work = 1/2 k x^2 = (1/2) * (1 kN / m) * (.02 m)^2

answer is in Joules

Answer 2

F = 2N/0.2cm * 2cm = 10 N.
Work = F * d = 10 * 0.02 = 0.2 Joules.

Answer 3

Henry

yes , work does equal (force * distance)

the problem with a spring, is that the force is ALSO related to the distance

that's why the distance is squared in the spring work equation

Answer 4

Thanks for the explanation and information; I've used the Eq.(F*d) many times but never in this manner. Thanks!