# Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 115: 47

a) 390 N b) 7.7 m/s, directed toward the second base

#### Work Step by Step

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

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