## Physics (10th Edition)

a) Frictional force = 0.49 $\times$ mg = 0.49 $\times$ 81 $\times$ 9.8 $\approx$ 390 N a = $\frac{F_{net}}{m}$ $\approx$ $\frac{-390}{81}$ $\approx$ -4.8 m/$s^{2}$ b) u = v - at u = 0 - (-4.8)(1.6) u $\approx$ 7.7 m/s, directed toward the second base