300 J of work is done in sliding a 2 kg block up an inclined plane of height 10m .the work done against friction is:
(A) zero
(B) 100 J
(C) 300J
(D)200J
Answers
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1
Answer:
(B) 100 J
Explanation:
Given,
Work done =300J, mass=2kg,height=10m,g=10m/s
2
we know that the work done =fd
300=(mgsinθ+f)×
sinθ
h
Work-done against the friction:
f.
sinθ
h
=300−mgsinθ×
sinθ
h
⇒300−mgh=300−2×10×10=100J
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