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Q.

A particle of mass 500 gm is moving in a straight line with velocity v=bx^5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is : (Take b=0.25 m−3/2s⁻¹).

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a

2 J

b

4 J

c

16 J

d

8 J

answer is D.

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Detailed Solution

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By work energy theorem
Work done by all forces $= Δ K.E.$
$\begin{array}{l} \Rightarrow W = \frac{1}{2} {mv}_{f}^{2} - \frac{1}{2} {mv}_{i}^{2} \\ \Rightarrow W = \frac{1}{2} \times 0 .5 \times (0 .25)^{2} \times (4)^{5} \\ \Rightarrow W = 16 J \end{array}$

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