Question

A body starts from rest and moves with uniform acceleration. What is the ratio of kinetic energies at the end of 1st, 2nd and 3rd seconds of its journey ​

Answer 1

Answer:

Ratio = 1:4:9

Explanation:

In the question,

Initial velocity of the body, u= 0 m/s

Acceleration, a = a m/s² (constant)

Also,

Kinetic energy is given by,

$K=\frac{1}{2}mv^{2}\\  where,\\v=u+at\\and,\\v=(0)+at\\v=at\\So,\\K=\frac{1}{2}m(at)^{2}=\frac{1}{2}ma^{2}   t^{2}$

So,

At the end of 1st second,

$K=\frac{1}{2}ma^{2}  (1)^{2}= \frac{1}{2}ma^{2}$

At the end of 2nd second,

$K=\frac{1}{2}ma^{2}  (2)^{2}= \frac{4}{2}ma^{2}$

Similarly, at the end of 3rd second,

$K=\frac{1}{2}ma^{2}  (3)^{2}= \frac{9}{2}ma^{2}$

So,

Ratio of Kinetic Energies at 1st, 2nd and 3rd second will be,

$\frac{\frac{\frac{1}{2}ma^{2}  }{\frac{4}{2}ma^{2}}}{\frac{9}{2}ma^{2}}  =1:4:9$