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Work Power and Energy question

2023 · 13 Apr · Shift 2 · Q71
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Work Power and Energy question

2023 · 13 Apr · Shift 2 · Q71

JEE MainPhysicsWork Power and EnergyNumerical+4 / −1
A car accelerates from rest to u m/su \mathrm{~m} / \mathrm{s}u m/s. The energy spent in this process is E J. The energy required to accelerate the car from u m/su \mathrm{~m} / \mathrm{s}u m/s to 2um/s2 \mathrm{u} \mathrm{m} / \mathrm{s}2um/s is nE J\mathrm{nE~J}nE J. The value of n\mathrm{n}n is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 3

  1. Use work-energy theorem

The energy spent in accelerating the car equals the change in kinetic energy.

Kinetic energy of a body of mass mmm moving with speed vvv is: K=12mv2K = \frac{1}{2}mv^2K=21​mv2

  1. From rest to uuu

Initial speed =0=0=0, final speed =u=u=u.

So the energy spent is E=12mu2−0=12mu2E = \frac{1}{2}mu^2 - 0 = \frac{1}{2}mu^2E=21​mu2−0=21​mu2

  1. From uuu to 2u2u2u

Required energy == = change in kinetic energy: ΔK=12m(2u)2−12mu2\Delta K = \frac{1}{2}m(2u)^2 - \frac{1}{2}mu^2ΔK=21​m(2u)2−21​mu2 =12m(4u2−u2)= \frac{1}{2}m(4u^2-u^2)=21​m(4u2−u2) =12m(3u2)= \frac{1}{2}m(3u^2)=21​m(3u2) =3(12mu2)= 3\left(\frac{1}{2}mu^2\right)=3(21​mu2)

But from step 2, 12mu2=E\frac{1}{2}mu^2 = E21​mu2=E

Hence, ΔK=3E\Delta K = 3EΔK=3E

So, n=3n=3n=3

  1. Comparison with stored answer

Derived answer is 333, which matches the stored correct answer.

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