## College Physics (4th Edition)

(a) The acceleration of the disk is $-0.853~m/s^2$ (b) The coefficient of kinetic friction is $0.087$
(a) We can find the acceleration: $v_f^2 = v_0^2+2ad$ $a = \frac{v_f^2 - v_0^2}{2d}$ $a = \frac{0 - (3.2~m/s)^2}{(2)(6.0~m)}$ $a = -0.853~m/s^2$ The acceleration of the disk is $-0.853~m/s^2$ (b) We can use the magnitude of the acceleration to find the coefficient of kinetic friction: $mg~\mu_k = ma$ $\mu_k = \frac{a}{g}$ $\mu_k = \frac{0.853~m/s^2}{9.80~m/s^2}$ $\mu_k = 0.087$ The coefficient of kinetic friction is $0.087$