Laws of MotionHard
Question
A block of mass m is at rest with respect to a lift when placed on inclined plane of inclination θ inside the lift. If lift move upward at constant velocity v then work done by friction force on block in t time is :-
Options
A.Zero
B.mgt2vcos2θ
C.mgtvsin2θ
D.
mgtv sin 2θ
mgtv sin 2θSolution
w = F d cos θ
w = f(vt) cos (90 - θ)
∵ Angle between friction and displacement is
(90o - θ)
∵ f = mgsin θ
∴ w = (mgsinθ) (vt)(sinθ)
= mgvtsin2θ
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