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Question

300 J of work is done in sliding a 2 kg block up the inclined plane of height 10 m with constant velocity. The work done against friction is (Take g=10 m/s2)

A
100 J
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B
Zero J
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C
200 J
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D
500 J
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Solution

The correct option is A 100 J
Let the applied force F be parallel to the inclined plane. When it pulls the block up, kinetic friction acts down to oppose the relative motion.

The work done by the force to displace the block through a distance l along the inclined plane is given as
W=F.s=Fl


l=h cosec θ
W=Fh cosec θ (2)
Since the block slides up with constant velocity, therefore the net force acting on the block
Fx=0; Fy=0mg sin θ+fk=F (3)

Using 2 and 3 we get
W=(mg sin θ+fk)h cosec θ

=mgh+fkh cosec θ

=300 J ......(given)

Wk(Work done against friction)=fkh cosec θ=300 Jmgh=300 J(2×10×10) J=300 J200 J=100 J

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