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Work Energy Power

JEE Main level, single correct, medium

19 of 183

Question 19 of 183

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

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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}=\Delta K

=\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

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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More Physics practice

JEE Main Physics practice question on Work Energy Power (Work done), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

A particle of mass 500g500g500g is moving in a straight line with velocity v=bx52v=b{x}^{\dfrac{5}{2}}v=bx25​. The work done by the net force during its displacement from x=0x=0x=0 to x=4mx=4mx=4m is (Take b=0.25m−32s−1b=0.25{m}^{\dfrac{-3}{2}}{s}^{-1}b=0.25m2−3​s−1).

JEE Main question paper PDF

More Physics practice

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