What frictional force must be overcome to pull a 50 kg sack of rice along a cement floor if the coefficient of friction is 0.22?
a. 50 N
b. 196 N
c. 490 N
d. 108 N
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We know frictional force f=μN where N=normal force=mg
Given sliding frictionμ=0.25
Thus, f=0.25×50×10=125N
Hence F
net
=F
app
−f=575−125=450N
Also we know ForceF=ma
a=
50
450
=9m/s
2
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