Question

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking $g=10\,m/{{s}^{2}},$ the work done against friction is

• A. 200 J
• B. 100 J
• C. zero
• D. 1000 J

Answer 1

Answer: (B)

100 J

Answer 2

Net work done in sliding a body up to a height h on inclined plane = Work done against gravitational force + Work done against frictional force $\Rightarrow$ $W={{W}_{g}}+{{W}_{g}}$ ?(i) but $W=300\,J$ ${{W}_{g}}=mgh=2\times 10\times 10=200\,J$ Putting in E (i) we get $300=200+{{W}_{f}}$ ${{W}_{f}}=300-200=100\,\text{J}$