Question

An object is acted upon by a constant force which gives it a constant acceleration a. At a certain time t1, having started from rest at t = 0, it has kinetic energy K1. At what time t2 has its kinetic energy doubled?

Answer 1

Answer:

$t_2=\sqrt{2}(t_1)$

Explanation:

The kinetic energy of a body is that energy it possesses due to its motion. In classical mechanics, this energy depends only on its mass and speed, as follows:

$K=\frac{mv^2}{2}$

The speed in an uniformly accelerated motion is given by:

$v=at$

Replacing this expression in the formula for the kinetic energy, we have:

$K=\frac{ma^2t^2}{2}$

So, if we have $K_2=2K_1$:

$K_1=\frac{ma^2t_1^2}{2}(1)\\K_2=\frac{ma^2t_2^2}{2}\\2K_1=\frac{ma^2t_2^2}{2}\\K_1=\frac{ma^2t_2^2}{4}(2)$

Equaling (1) and (2) and solving for $t_2$:

$\frac{ma^2t_1^2}{2}=\frac{ma^2t_2^2}{4}\\t_2=\frac{4t_1^2}{2}\\t_2=\sqrt{2t_1^2}\\t_2=\sqrt{2}(t_1)$