Question

A 68.1 kg runner has a speed of 3.90 m/s at one instant during a long-distance event.HINT=(a) Apply the definition of kinetic energy. (b) Use the work-energy theorem.(a) What is the runner's kinetic energy at this instant (in J)?_____ J(b) How much net work (in J) is required to double his speed?_____ J

Answer 1

We will have the following:

a)

$\begin{gathered} k=\frac{1}{2}mv^2\Rightarrow k=\frac{1}{2}(68.1kg)(3.90m/s)^2 \\ \\ \Rightarrow k=517.9005J\Rightarrow k\approx517.90J \end{gathered}$

So, the energy is approximately 517.90 Joules.

b.

$\begin{gathered} W=\frac{1}{2}(68.1kg)(2\ast3.90m/s)^2-\frac{1}{2}(68.1kg)(3.90m/s)^2\Rightarrow W=2466.3015J \\ \\ \Rightarrow W\approx2466.30J \end{gathered}$

And the work required to double its speed is approximatley 2466.30 Joules.