Q123 / 200 · JIPMER 2008 · Physics

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

  • A
    $200 J$
  • Correct answer
    B
    $100 J$
  • C
    $zero$
  • D
    $1000 J$

Solution

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

Hence, the correct answer is option: 'B'.

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