## College Physics (4th Edition)

The kinetic energy decreases by $0.03125~J$
Let $d_e$ be the distance the spring is stretched at the equilibrium position. We can find $d_e$: $k~d_e = mg$ $d_e = \frac{mg}{k}$ $d_e = \frac{(0.50~kg)(9.80~m/s^2)}{25~N/m}$ $d_e = 0.196~m$ We can find the spring potential energy at the equilibrium position: $U_s = \frac{1}{2}kd_e^2$ $U_s = \frac{1}{2}(25~N/m)(0.196~m)^2$ $U_s = 0.4802~J$ We can find the spring potential energy at a position 5.0 cm lower than the equilibrium position: $U_s = \frac{1}{2}k(d_e+d)^2$ $U_s = \frac{1}{2}(25~N/m)(0.196~m+0.050~m)^2$ $U_s = 0.75645~J$ We can find the change in spring potential energy between these two positions: $\Delta U_s = 0.75645~J-0.4802~J$ $\Delta U_s = 0.27625~J$ The total energy in the system is constant. Therefore the change in total energy in the system from one position to another position is zero. We can find the change in kinetic energy: $\Delta K + \Delta U_g + \Delta U_s = 0$ $\Delta K = -\Delta U_g - \Delta U_s$ $\Delta K = -(mg\Delta h) - \Delta U_s$ $\Delta K = -(0.50~kg)(9.80~m/s^2)(-0.050~m) - (0.27625~J)$ $\Delta K = -0.03125~J$ The kinetic energy decreases by $0.03125~J$