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

(a) $v = 3.94~m/s$ (b) The magnitude of the acceleration is $11.8~m/s^2$ and the acceleration is directed away from the post.
(a) The kinetic energy of the block will be equal to the sum of the work done by the force and the spring. $\frac{1}{2}mv^2 = F~x - \frac{1}{2}kx^2$ $v^2 = \frac{2F~x - kx^2}{m}$ $v = \sqrt{\frac{2F~x - kx^2}{m}}$ $v = \sqrt{\frac{(2)(54.0~N)(0.400~m) - (76.0~N/m)(0.400~m)^2}{2.00~kg}}$ $v = 3.94~m/s$ (b) We can use a force equation to find the acceleration. $\sum F = ma$ $F-kx = ma$ $a = \frac{F-kx}{m}$ $a = \frac{54.0~N-(76.0~N/m)(0.400~m)}{2.00~kg}$ $a = 11.8~m/s^2$ The magnitude of the acceleration is $11.8~m/s^2$ and the acceleration is directed away from the post.