1.

300 J of work is done in sliding a 2 kg block up on inclined plane of height 10m. Work done against frictin is (g=10ms^(-2))

Answer»

1000 j
200 J
100 J
Zero

Solution : TOTAL work DONE =GAIN in P.E. + Work done against friction
`300=2xx10xx10+WimpliesW=100J`


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