JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 \(b =0.25 \,m ^{-3 / 2} s ^{-1}\) )
- A \(2\)
- B \(4\)
- C \(8\)
- D \(16\)
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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