Question

The force an ideal spring exerts on an object is given by Fx = -kx, where x measures the displacement of the object from its equilibrium (x = 0) position. If k = 60 N/m, how much work is done by this force as the object moves from x = -0.20 m to x = 0?

Answer 1

Answer:

The work done by this force can be found via the following formula

$W = \int{F(x)} \, dx = \int\limits^0_{-20} {(-kx)} \, dx = \frac{-kx^2}{2}\left \{ {{x=0} \atop {x=-20}} \right. = \frac{-60*(-20)^2}{2} \\W = -12000J$

Explanation:

Alternatively, the work done by the object is equal to the elastic potantial energy done by the spring.

$U = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 =0 - \frac{1}{2}60(-20)^2 = -12000J$