## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We can find the force $F_s$ that the spring pushes up on the student. $\sum F = ma$ $F_s-mg = ma$ $F_s = m(g+a)$ $F_s = (60~kg)(9.80~m/s^2+3.0~m/s^2)$ $F_s = 768~N$ We can find the distance that the spring is compressed when the spring is pushing with this force. $kx = F_s$ $x = \frac{F_s}{k}$ $x = \frac{768~N}{2.5\times 10^3~N/m}$ $x = 0.31~m$ The spring is compressed by 0.31 meters.