300J of work is done in sliding a 2kg block up the inclined plane of height 10m with constant velocity. The work done against friction is (Take g=10m/s2)
A
100J
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B
Zero J
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C
200J
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D
500J
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Solution
The correct option is A100J
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=hcosecθ ⇒W=Fhcosecθ……(2)
Since the block slides up with constant velocity, therefore the net force acting on the block ∑Fx=0;∑Fy=0⇒mgsinθ+fk=F……(3)
Using 2 and 3 we get W=(mgsinθ+fk)hcosecθ
=mgh+fkh cosecθ
=300 J ......(given)
∴Wk(Work done against friction)=fkh cosecθ=300 J−mgh=300J−(2×10×10)J=300 J−200J=100 J