JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2022
- Medium
- Numerical
A particle of mass \(500 \,gm\) is moving in a straight line with velocity \(v=b x^{5 / 2}\). The work done by the net force during its displacement from \(x=0\) to \(x =4 \,m\) is ...................\(J\) (Take \(b =0.25 \,m ^{-3 / 2} s ^{-1}\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(16\)
Step-by-step Solution
Detailed explanation
By work energy theorem work done by net force \(=\Delta K.E.\)
\(\Rightarrow w =\frac{1}{2} mu _{ f }^{2}-\frac{1}{2} mv _{ i }^{2}\)
\(w =\frac{1}{2} \times 0.5 \times(0.25)^{2} \times(4)^{5}\)
\(W =16 \,J\)
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