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:

  1. (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 $=300-200$ $=100 J$



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