# A 60.6 kg runner has a speed of 3.10 m/s at one instant during a long-distance event. (a) What is...

## Question:

A 60.6 kg runner has a speed of 3.10 m/s at one instant during a long-distance event.

(a) What is the runner's kinetic energy at this instant (in J)?

(b) How much net work (in J) is required to double her speed?

## Kinetic Energy:

The quantity known as the kinetic energy is the energy associated with the movement of a body. We can acquire the kinetic energy by applying the equation, {eq}\displaystyle KE = \frac{1}{2}mv^2 {/eq}, where m is the mass and v is the velocity of the object.

Determine the kinetic energy of the runner by applying the equation, {eq}\displaystyle KE = \frac{1}{2}mv^2 {/eq}. We are given that the mass of the runner is {eq}\displaystyle m = 60.6\ kg {/eq}, and that his velocity is {eq}\displaystyle v = 3.10\ m/s {/eq}. We simply plug in the given values to determine the answer.

{eq}\begin{align} \displaystyle KE &= \frac{1}{2}mv^2\\ &= \frac{1}{2}(60.0\ kg)(3.10\ m/s)^2\\ &\approx\boxed{\rm 288\ J} \end{align} {/eq}

Now, we acquire the net work, W, by finding the difference in the kinetic energies as the runner's speed is doubled. We determine the answer.

{eq}\begin{align} \displaystyle W &= KE_1 - KE_2\\ &= \frac{1}{2}m(2v)^2 - \frac{1}{2}mv^2 &= \frac{1}{2}(60.6\ kg)(2(3.10\ m/s))^2 - \frac{1}{2}(60.6\ kg)(3.10\ m/s)^2\\ &\approx\boxed{\rm 874\ J} \end{align} {/eq}