JEE MAIN - Physics (2023 - 13th April Evening Shift - No. 28)
A car accelerates from rest to $$u \mathrm{~m} / \mathrm{s}$$. The energy spent in this process is E J. The energy required to accelerate the car from $$u \mathrm{~m} / \mathrm{s}$$ to $$2 \mathrm{u} \mathrm{m} / \mathrm{s}$$ is $$\mathrm{nE~J}$$. The value of $$\mathrm{n}$$ is ____________.
Answer
3
Explanation
The kinetic energy of a moving object of mass $$m$$ and velocity $$v$$ is given by the formula:
$$K = \frac{1}{2}mv^2$$
The work done in accelerating an object from rest to velocity $$v$$ is equal to its change in kinetic energy. Therefore, the energy spent in accelerating the car from rest to $$u \mathrm{~m}/\mathrm{s}$$ is:
$$E = \frac{1}{2}mu^2$$
The energy required to accelerate the car from $$u \mathrm{~m}/\mathrm{s}$$ to $$2u \mathrm{~m}/\mathrm{s}$$ is:
$$\begin{aligned} nE &= \frac{1}{2}m(2u)^2 - \frac{1}{2}mu^2 \\\\ &= 2mu^2 - \frac{1}{2}mu^2 \\\\ &= \frac{3}{2}mu^2 \\\\ &= 3E \end{aligned} $$
$$ \therefore $$ n = 3
$$K = \frac{1}{2}mv^2$$
The work done in accelerating an object from rest to velocity $$v$$ is equal to its change in kinetic energy. Therefore, the energy spent in accelerating the car from rest to $$u \mathrm{~m}/\mathrm{s}$$ is:
$$E = \frac{1}{2}mu^2$$
The energy required to accelerate the car from $$u \mathrm{~m}/\mathrm{s}$$ to $$2u \mathrm{~m}/\mathrm{s}$$ is:
$$\begin{aligned} nE &= \frac{1}{2}m(2u)^2 - \frac{1}{2}mu^2 \\\\ &= 2mu^2 - \frac{1}{2}mu^2 \\\\ &= \frac{3}{2}mu^2 \\\\ &= 3E \end{aligned} $$
$$ \therefore $$ n = 3
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