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