Q104 / 200 · NEET 2009 (Screening) · Physics

A body of mass $1 \mathrm{~kg}$ is thrown upwards with a velocity $20 \mathrm{~ms}^{-1}$. It momentarily comes to rest after attaining a height of $18 \mathrm{~m}$. How much energy is lost due to air friction ? $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$

  • Correct answer
    A
    $20 \mathrm{~J}$
  • B
    $30 \mathrm{~J}$
  • C
    $40 \mathrm{~J}$
  • D
    $10 \mathrm{~J}$

Solution

Key Idea The energy lost due to air friction is equal to difference of initial kinetic energy and final potential energy. Initially body posses only kinetic energy and after attaining a height the kinetic energy is zero. Therefore, loss of energy $=\mathrm{KE}-\mathrm{PE}$ $$ \begin{aligned} & =\frac{1}{2} \mathrm{mv}^2-\mathrm{mgh} \\ & =\frac{1}{2} \times 1 \times 400-1 \times 18 \times 10 \\ & =200-180=20 \mathrm{~J} \end{aligned} $$

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

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