300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. The work done against friction is (Take g = 10m/s2)
A
zero
B
100J
C
200J
D
300J
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Solution
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Given,
Work done =300J, mass=2kg,height=10m,g=10m/s2
we know that the work done =fd
300=(mgsinθ+f)×hsinθ
Work-done against the friction:
f.hsinθ=300−mgsinθ×hsinθ⇒300−mgh=300−2×10×10=100J
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