Question

an athlete with a mass of 65.0 kg is in a cross country race, and at one instant he has a speed of 3.30 m/s. (a) what is his kinetic energy at this instant? (ke_{i}=) j (b) if he wishes to doubles his speed to reach the finish line, his kinetic energy will change by what factor? (\frac{ke_{f}}{ke_{i}}=)

Explanation:

Step1: Recall kinetic - energy formula

The formula for kinetic energy is $KE=\frac{1}{2}mv^{2}$, where $m$ is the mass and $v$ is the speed.

Step2: Calculate initial kinetic energy for part (a)

Given $m = 65.0\ kg$ and $v = 3.30\ m/s$. Substitute into the formula:
$KE_{i}=\frac{1}{2}\times65.0\times(3.30)^{2}=\frac{1}{2}\times65.0\times10.89 = 353.925\ J$

Step3: Analyze the effect of speed - doubling on kinetic energy for part (b)

Let the initial speed be $v_{i}$ and the final speed be $v_{f}=2v_{i}$. The initial kinetic energy is $KE_{i}=\frac{1}{2}mv_{i}^{2}$, and the final kinetic energy is $KE_{f}=\frac{1}{2}mv_{f}^{2}=\frac{1}{2}m(2v_{i})^{2}=\frac{1}{2}m\times4v_{i}^{2}=4\times\frac{1}{2}mv_{i}^{2}$.
So, $\frac{KE_{f}}{KE_{i}} = 4$.

Answer:

(a) $353.925$
(b) $4$