A body of mass $2 \mathrm{~kg}$ thrown vertically from the ground with a velocity of $8 \mathrm{~ms}^{-1}$…

A body of mass $2 \mathrm{~kg}$ thrown vertically from the ground with a velocity of $8 \mathrm{~ms}^{-1}$ reaches a maximum height of $3 \mathrm{~m}$. The work done by the air resistance is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
  1. 4J
  2. 60J
  3. 64J
  4. 8J

Solution

Initial energy of particle $ =\frac{1}{2} m v^2=\frac{1}{2} \times 2 \times 8^2=64 \mathrm{~J} $ Final energy of particle at maximum height $ =m g h=2 \times 10 \times 3=60 \mathrm{~J} $ Work done against air friction $ =\text { Loss of energy }=64-60=4 \mathrm{~J} $

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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