question_answer

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. The work done against friction is \[\left( \text{Take g}=\text{ 1}0\text{ m}/{{\text{s}}^{\text{2}}} \right)\]

A)  zero                                     

B)  100 J

C)  200 J                                    

D)  300 J

Correct Answer: B

Solution :

                 Work done against friction is equal to                 W = W - U                 where U is potential energy, W the work done in sliding the block up the inclined plane.                 U = mgh                 \[=2\times 10\times 10=200J\]                 W = 300 J                 W = 300 - 200 = 100 J