AP EAMCET · PHYSICS · Work Power Energy
A bullet of mass \(10 \mathrm{~g}\) is fired horizontally with a velocity \(1000 \mathrm{~ms}^{-1}\) from a rifle situated at a height \(50 \mathrm{~m}\) above the ground. If the bullet reaches the ground with a velocity \(500 \mathrm{~ms}^{-1}\), the work done against air resistance in the trajectory of the bullet is : \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
- A \(5005 \mathrm{~J}\)
- B \(3755 \mathrm{~J}\)
- C \(3750 \mathrm{~J}\)
- D \(17.5 \mathrm{~J}\)
Answer & Solution
Correct Answer
(C) \(3750 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
From equation of motion, \(v^2=u^2-2 a s\) \((500)^2=(1000)^2-2 \times a \times s\) \(s=\frac{(1000)^2-(500)^2}{2 a}=\frac{375000}{a}\) \(\therefore\) Work done against air resistance \(=F S\) \(=m a \times s\) \(=\frac{10}{1000} a \times \frac{375000}{a}\) \(=3750 \mathrm{~J}\)
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