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

We know the total energy $E$ is equal to the sum of the potential energy $U$ and the kinetic energy $K$. At point A, the particle is at rest. We can find the total energy. $E = U+K$ $E = 5.0~J+0$ $E = 5.0~J$ The total energy is 5.0 J We can find the kinetic energy at point B; $U+K = E$ $U+K = 5.0~J$ $K = 5.0~J - U$ $K = 5.0~J - 2.0~J$ $K = 3.0~J$ We can find the speed of the particle at point B; $K = 3.0~J$ $\frac{1}{2}mv^2 = 3.0~J$ $v^2 = \frac{(2)(3.0~J)}{m}$ $v = \sqrt{\frac{(2)(3.0~J)}{m}}$ $v = \sqrt{\frac{(2)(3.0~J)}{0.50~kg}}$ $v = 3.5~m/s$ We can find the kinetic energy at point C; $U+K = E$ $U+K = 5.0~J$ $K = 5.0~J - U$ $K = 5.0~J - 3.0~J$ $K = 2.0~J$ We can find the speed of the particle at point C; $K = 2.0~J$ $\frac{1}{2}mv^2 = 2.0~J$ $v^2 = \frac{(2)(2.0~J)}{m}$ $v = \sqrt{\frac{(2)(2.0~J)}{m}}$ $v = \sqrt{\frac{(2)(2.0~J)}{0.50~kg}}$ $v = 2.8~m/s$ We can find the kinetic energy at point D; $U+K = E$ $U+K = 5.0~J$ $K = 5.0~J - U$ $K = 5.0~J - 0$ $K = 5.0~J$ We can find the speed of the particle at point D; $K = 5.0~J$ $\frac{1}{2}mv^2 = 5.0~J$ $v^2 = \frac{(2)(5.0~J)}{m}$ $v = \sqrt{\frac{(2)(5.0~J)}{m}}$ $v = \sqrt{\frac{(2)(5.0~J)}{0.50~kg}}$ $v = 4.5~m/s$