Work Energy Power

JEE Main level, numerical answer, medium

153 of 183

Question 153 of 183

A body of mass 2kg2kg is initially at rest. It starts moving unidirectionally under the influence of a source of constant power PP. Its displacement in 4s4s is 13α2Pm\dfrac{1}{3}{\alpha }^{2}\sqrt{P}m. The value of α\alpha will be ______.

Show answer and solution

Answer: 4

Energy provided will be used to increase kinetic energy of the body. Therefore,
12mV2=Pt\dfrac{1}{2}m{V}^{2}=Pt
V=2PtmV=\sqrt{\dfrac{2Pt}{m}}
⇒dxdt=2Ptm⇒∫0xdx=2Pm∫04t12dt\Rightarrow \dfrac{dx}{dt}=\sqrt{\dfrac{2Pt}{m}} \Rightarrow {\int }_{0}^{x}dx=\sqrt{\dfrac{2P}{m}}{\int }_{0}^{4}{t}^{\dfrac{1}{2}}dt
⇒x=2Pm23[t32]04\Rightarrow x=\sqrt{\dfrac{2P}{m}}\dfrac{2}{3}{\left[{t}^{\dfrac{3}{2}}\right]}_{0}^{4}
⇒x=16P3=13×16P\Rightarrow x=\dfrac{16\sqrt{P}}{3}=\dfrac{1}{3}\times 16\sqrt{P}
Hence, required value of α=4\alpha =4.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Collisions, centre of mass and rockets simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

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