Question

A 400-g block with an initial speed of 80 cm/s slides along a horizontal tabletop against a friction force of 0.70 N. (a) How far will it slide before stopping? (b) What is the coefficient of friction between the block and the tabletop?

Answer 1

Given:

$m=400\:\rm g=0.4\: kg$

$v=80\:\rm cm/s=0.8\: m/s$

$F_f=0.70\:\rm N$

(a) The energy-work theorem says

$$0-\frac{mv^2}{2}=W=-F_fd$$

Hence

$$d=\frac{mv^2}{2F}=\frac{0.4*0.8^2}{2*0.7}=0.18\:\rm m$$

(b) The friction force

$$F_f=\mu N=\mu mg$$

Hence

$$\mu=\frac{F_f}{mg}=\frac{0.7}{0.4*9.8}=0.18$$