Question

A 55 kg softball player slides into second base, generating 940 J of thermal energy in her legs and the ground. How fast was she running?

Answer 1

When the player slides into the base, all its kinetic energy converts into thermal energy in the legs and the ground.
The thermal energy generated is 940 J, so the initial kinetic energy of the player must be 940 J as well:
$K= \frac{1}{2}mv^2=940 J$
where m=55 kg is the mass of the player and v is his speed. Re-arranging the fomula, we can find the value of his speed, v:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 940 J}{55 kg} } =7.4 m/s$