A steel block s is placed on a platform b. The weight of the block is 5N.A steadily increasing is applied horizontally to the block so that it just begin to move. If the coefficient of the friction is 0.2,calculate the minimum force required to move the block
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a. F
fric
=μ
s
W=0.4∗20=8N
b. Since F
applied
<F
fric
F
fric
=5N
c. Minimum force required to start the motion F=F
staticfric
=8N
d. Minimum force required to start the motion
F=F
kineticfric
=μ
k
W=0.2∗20=4N
e. F>F
staticfric
motion started.
F>F
kineticfric
hence motion continues.
F
fric
=4N
Answered By=shreyash m.............
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