Answer
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
