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  4. 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

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Q. $300\, J$ 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

JIPMERJIPMER 2011Work, Energy and Power

Solution:

Loss in potential energy $=m g h$
$=2 \times 10 \times 10$
$=200\, J$
Gain in kinetic energy $=$ work done $=300 \,J$
$\therefore $ Work done against friction $=300-200=100\, J$