Question

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

Answer 1

Answer:

Given:

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

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

F_f=0.70\:\rm NFf=0.70N

(a) The energy-work theorem says

0-\frac{mv^2}{2}=W=-F_fd0−2mv2=W=−Ffd

Hence

d=\frac{mv^2}{2F}=\frac{0.4*0.8^2}{2*0.7}=0.18\:\rm md=2Fmv2=2∗0.70.4∗0.82=0.18m

(b) The friction force

F_f=\mu N=\mu mgFf=μN=μmg

Hence

\mu=\frac{F_f}{mg}=\frac{0.7}{0.4*9.8}=0.18μ=mgFf=0.4∗9.80.7=0.18