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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 \(b =0.25 \,m ^{-3 / 2} s ^{-1}\) )

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Correct Answer

(D) \(16\)

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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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