Work Energy and Power - Result Question 7
9. $300 J$ of work is done in sliding a $2 kg$ block up an inclined plane of height $10 m$. Taking $g=10$ $m / s^{2}$, work done against friction is
[2006]
(a) $100 J$
(b) zero
(c) $1000 J$
(d) $200 J$
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Answer:
Correct Answer: 9. (a)
Solution:
- (a) Work done against gravity $=m g \sin \theta \times d$ $(d \sin \theta=10)$
$ =2 \times 10 \times 10 $
$ =200 J $
Actual work done $=300 J$
Work done against friction $=300-200$ $=100 J$
