1
Question
A small block of mass m is kept on a rough inclined plane of angle Θ fixed in an elevator. The elevator uniform velocity v and the block does not slide down. What is the work done by the frictional force on the block.
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Solution
F=mg sinθ now break f
into perpendicular components.
here fsinθ will act in
downward drtn. and f cosθ in a direction perpendicular to motion.
so the work will be
done by f sinθ.
work done=mgsin2θ *
h
h = height travelled
in time t
h = v t
Hence
Work done = mg vt
sin²θ
F=mg sinθ now break f into perpendicular components.
here fsinθ will act in downward drtn. and f cosθ in a direction perpendicular to motion.
so the work will be done by f sinθ.
work done=mgsin2θ * h
h = height travelled in time t
h = v t
Hence
Work done = mg vt sin²θ
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