A particle of mass 500g 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 = 4m is (Take b = 0 . 25m(- 3)/(2) s ^ - 1 ).

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Correct answer: D. 16J

According to the work-energy theorem, work done by all forces will be is equal to the change in kinetic energy.

${W}_{all} = \DeltaK$

$= \dfrac{1}{2}m{b}^{2}\left({x}_{2}^{5} - {x}_{1}^{5}\right)$ = \dfrac{1}{2} \times 0 . 5 \times \dfrac{1}{16} \times {4}^{5} = 16J

A particle of mass 500g is moving in a straight line with velocity...

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