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Question

# 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)zero100J200J300J

A
100J
B
200J
C
zero
D
300J
Solution
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#### Given,Work done =300J, mass=2kg,height=10m,g=10m/s2we know that the work done =fd300=(mgsinθ+f)×hsinθWork-done against the friction:f.hsinθ=300−mgsinθ×hsinθ⇒300−mgh=300−2×10×10=100J

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