## College Physics (4th Edition)

(a) The kinetic energy of the sack is $0.70~J$ (b) The speed of the sack is $0.37~m/s$
(a) We can find the work that Sam does on the sack: $Work = F~d$ $Work = (2.0~N)(0.35~m)$ $Work = 0.70~J$ The work that Sam does on the sack is equal to the change in kinetic energy of the sack. Therefore, the kinetic energy of the sack is $0.70~J$ (b) We can find the speed of the sack: $KE = 0.70~J$ $\frac{1}{2}mv^2 = 0.70~J$ $v = \sqrt{\frac{(2)(0.70~J)}{m}}$ $v = \sqrt{\frac{(2)(0.70~J)}{10.0~kg}}$ $v = 0.37~m/s$ The speed of the sack is $0.37~m/s$