Questions
A block B of mass 2 kg is placed on the rough inclined surface of a wedge W. The wedge is kept on a smooth horizontal floor and held in position by an external agent. The block is in equilibrium on the wedge. If the wedge slowly moves to the left by 5m,
a
Work done by force of friction on the block is -203J
b
Work done by normal force acting on the block is +253J
c
Work done by force of friction on the block is +253J
d
None of the above is correct.
detailed solution
Correct option is B
f=mgsinθ and N=mgcosθ∴ f=2×10×sin30oN=10 NN=2×10×cos30oN=103 N∴ Wf=−fcosθS=−10×cos30o×5J=−253 JWN=Nsinθ×S=103×sin30o×5 J=+253 JTalk to our academic expert!
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