## College Physics (4th Edition)

The maximum kinetic energy of the body is $4.0\times 10^{-6}~J$
In general: $y(t) = A~sin(\omega~t+\phi)$ In this case: $y(t) = (4.0~cm)~sin~[~(0.70~rad/s)~t~]$ We can see that $~A = 4.0~cm~$ and $~\omega = 0.70~rad/s~$ At the equilibrium position, the upward force of the spring is equal in magnitude to the weight. We can find the mass of the body: $mg = kd$ $m = \frac{kd}{g}$ $m = \frac{(2.5~N/m)(0.040~m)}{9.80~m/s^2}$ $m = 0.0102~kg$ We can find the maximum kinetic energy: $K_m = \frac{1}{2}mv_m^2$ $K_m = \frac{1}{2}m(A~\omega)^2$ $K_m = \frac{1}{2}(0.0102~kg)~(0.040~m)^2~(0.70~rad/s)^2$ $K_m = 4.0\times 10^{-6}~J$ The maximum kinetic energy of the body is $4.0\times 10^{-6}~J$.