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JIPMER 2008
Exam
JIPMER
Year
2008
Subject
Physics
Chapter
Work, Power and Energy

JIPMER 2008 · Physics · Work, Power and Energy

300 J of work is done in sliding a 2 kg block up an inclined plane...

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m . Taking g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^2, work done against friction is

Options

  1. A200J200 J
  2. B100J100 J
  3. Czerozero
  4. D1000J1000 J

Correct answer

B. 100J100 J

Step-by-step solution

Net work done in sliding a body up to a height hh on inclined plane = Work done against gravitational force + Work done against frictional force \begin{array}{r}=\text{ Work done against gravitational force } \\ \quad+\text{ Work done against frictional force }\end{array} \Rightarrow \quad W=W_g+W_f...(i)but...(i) butW=300  J\mathrm{~J} W_g=m g h=2 \times 10 \times 10=200  J\mathrm{~J} Putting in Eq. (i), we get 300=200+Wf⇒Wf=300−200=100 J\begin{gathered} \\ 300=200+W_f \\ \Rightarrow \quad W_f=300-200=100 \mathrm{~J}\end{gathered}
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