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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

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We will have the following:

a)


\begin{gathered} k=(1)/(2)mv^2\Rightarrow k=(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=(1)/(2)(68.1kg)(2\ast3.90m/s)^2-(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.

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