AP EAMCET · PHYSICS · Work Power Energy
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 \(=10 \mathrm{~ms}^{-2}\) )
- A 4J
- B 60J
- C 64J
- D 8J
Answer & Solution
Correct Answer
(A) 4J
Step-by-step Solution
Detailed explanation
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 \[ =\text { Loss of energy }=64-60=4 \mathrm{~J} \]
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