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  • A particle of mass is moving in a straight line with velocity <img alt="v=\mathrm{b} x^{5 / 2}" src="/latex-image/?v%3D%5Cm

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

Option: 1

2 \mathrm{~J}
 


Option: 2

4 \mathrm{~J}
 


Option: 3

8 \mathrm{~J}
 


Option: 4

16 \mathrm{~J}


Answers (1)

best_answer

\mathrm{v=bx^{5/2}}

Work energy theorem

\begin{aligned} \mathrm{W =\Delta K E } \\ \end{aligned}

\begin{aligned} \mathrm{=\frac{1}{2} m V_{2}^{2}-\frac{1}{2} m V_{1}^{2}} \\ \end{aligned}

\mathrm{=\frac{m}{2}\left[v_{2}^{2}-v_{1}^{2}\right]}

\mathrm{=\frac{0.5}{2}\times\left [ b^{2}\times(4^{5/2})^{2}-0 \right ]}

\mathrm{=\frac{0.5}{2} \times 0.25 \times 0.25 \times 4^{5}}

\mathrm{=16 J}

Hence, Correct answer is Option (2).

Posted by

Ritika Harsh

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