Question

A 67-kg base runner begins his slide into second base when he is moving at a speed of 3.6 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base. (a) How much mechanical energy is lost due to friction acting on the runner? J

Answer 1

Answer:

434.16 Joules

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of person

From work-energy theorem

$KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}\times 67\times (3.6^2-0)\\\Rightarrow KE=434.16\ Joules$

The runner loses 434.16 Joules of mechanical energy.

$W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow \mu mg=\frac{W}{s}\\\Rightarrow s=\frac{W}{\mu mg}\\\Rightarrow s=\frac{434.16}{0.7\times 67\times 9.81}\\\Rightarrow s=0.94364\ m$

He slides 0.94364 m