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)

A
zero
B
100J
C
200J
D
300J
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=300mgsinθ×hsinθ300mgh=3002×10×10=100J

983668_358825_ans_c934d73d8e7741cebb09b3ea74ad7540.PNG

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