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JEE Main 20238 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2023 · 8 April · Evening Shift · Physics · Work, Power and Energy

A body of mass 5kg is moving with a momentum of 10kgm s ^ - 1 . Now...

A body of mass 5kg5kg is moving with a momentum of 10kgms−110kgm{s}^{- 1}. Now a force of 2N2N acts on the body in the direction of its motion for 5s5s. The increase in the Kinetic energy of the body is _____ JJ.

Correct answer

0

Step-by-step solution

The formula to calculate the change on momentum of the object is given by

pf−pi=∫Fdt...(1)</p><p>Substitute{p}_{f} - {p}_{i} = ∫ Fdt . . . \left(1\right)</p><p>Substitute the values of the known parameters into equation (1) and solve to calculate the final momentum of the object.

{pf−10kgms−1=2N×5s⇒pf=(10+10)kgms−1=20kgms−1</p><p>Theformulato\begin{cases}{p}_{f} - 10kgm{s}^{- 1} & = & 2N \times 5s \\ & ⇒ & {p}_{f} = \left(10 + 10\right)kgm{s}^{- 1} \\ & = & 20kgm{s}^{- 1}\end{cases}</p><p>The formula tocalculate increase in kinetic energy \left(∆ K\right)oftheparticleisgivenbyof the particle is given by

∆K=∆ K =\dfrac{1}{2m}\left({{p}_{f}}^{2} - {{p}_{i}}^{2}\right) . . . \left(2\right)

Substitutethevaluesoftheknownparametersintoequation(2)tocalculatetherequiredincreaseinkineticenergy.</p><p>the values of the known parameters into equation (2) to calculate the required increase in kinetic energy.</p><p>\begin{cases}∆ K & = & \dfrac{1}{2 \times 5kg} \times \left({20}^{2} - {10}^{2}\right){kg}^{2}{m}^{2}{s}^{- 2} \\ & = & 30J\end{cases}

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