Question

A $55$ kg softball player slides into second base, generating $950$ J of thermal energy in her legs and the ground. How fast was she running?

Short answer

Speed of the player is $5.87$m/s

Step-by-step solution

Step 1. Kinetic to Thermal energy 

The frictional force is generated because of sliding; it's a dissipative force that converts themechanical energy to thermal energy.
$\mathrm{\Delta }K=-\mathrm{\Delta }{E}_{th}$
Here $\mathrm{\Delta }{E}_{th}$ is that the thermal energy generated and $\mathrm{\Delta K}$is change inmechanical energy.

Step 2. Find the velocity

Consider the above formula,
$\begin{array}{r}\mathrm{\Delta }K=-\mathrm{\Delta }{E}_{\text{th}}\end{array}$

$\begin{array}{r}\frac{1}{2}m\left(v_{\mathrm{f}}^2-v_i^2\right)=-\mathrm{\Delta }{E}_{\text{th}}\\ \end{array}$

$\left(v_{\mathrm{f}}^2-v_{\mathrm{i}}^2\right)=\frac{-2\mathrm{\Delta }{E}_{\text{th}}}{m}$

Here $v_{\mathrm{i}}^{}$ is that the initial speed, $v_f^{}$ is that the final speed and $m$ is that the mass of player.
Substitute $0\mathrm{m}/\mathrm{s}$for $v_{\mathrm{f}}.950\mathrm{J}$for$\mathrm{\Delta }{E}_{\mathrm{a}}$and $55\mathrm{kg}$for $m$

$\begin{array}{r}{v}_i=\sqrt{\frac{2\left(950\mathrm{J}\right)}{55\mathrm{kg}}}\\ \end{array}$

$=5.9\mathrm{m}/\mathrm{s}$

Therefore, the initial speed of the player was $=5.9\mathrm{m}/\mathrm{s}$.