## University Physics with Modern Physics (14th Edition)

(a) The force constant of the spring is $4.44\times 10^5~N/m$. (b) The spring must be compressed a distance of 0.128 meters.
We can set up two equations to solve this question. We can set up an energy equation. The kinetic energy of the satellite will be equal to the work done by the spring. $\frac{1}{2}kx^2 = \frac{1}{2}mv^2$ We can set up a force equation. Let's assume that the acceleration is 5.00g when the spring is released from its maximum compression. $kx = ma = 5.00~mg$ To solve for the distance $x$ that the spring is compressed, we can divide the energy equation by the force equation. $\frac{x}{2} = \frac{v^2}{10.0~g}$ $x = \frac{v^2}{5.00~g}$ $x = \frac{(2.50~m/s)^2}{(5.00)(9.80~m/s^2)}$ $x = 0.128~m$ We can use the distance of compression $x$ to find the force constant of the spring. $kx = 5.00~mg$ $k = \frac{5.00~mg}{x}$ $k = \frac{(5.00)(1160~kg)(9.80~m/s^2)}{0.128~m}$ $k = 4.44\times 10^5~N/m$ (a) The force constant of the spring is $4.44\times 10^5~N/m$. (b) The spring must be compressed a distance of 0.128 meters.