If a 70 kgbaseball player steals home by sliding into the plate with an initial speed of 10 m/sjust as he hits the ground, (a) what is the decrease in the player’s kinetic energy and (b) what is the increase in the thermal energy of his body and the ground along which he slides?

Mass of the baseball player, m = 70 kg

Initial speed of the baseball player, v = 10 m/s

Final speed of the baseball player, v = 0 m/s

Using the initial and final velocity of the player, find the decrease in the player’s kinetic energy. Then using the law of conservation of energy, find the increase in the thermal energy of his body and the ground along which he slides.

Formulae:

The kinetic energy is,

$KE=\frac{1}{2}m{v}^2$

The energy, E = constant

A decrease in the player’s kinetic energy is,

$$\begin{gathered} ∆KE=K{E}_i-K{E}_f \\ ∆KE=\frac{1}{2}m{v}_i^2-\frac{1}{2}m{v}_f^2 \\ =\frac{1}{2}m{v}_i^2-0 \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} ∆KE=\frac{1}{2}\left(70\right){\left(10\right)}^2 \\ =3500\mathrm{J} \\ =3.5\mathrm{KJ} \end{gathered}$$

Therefore, decrease in the player’s kinetic energy is 3.5 KJ .

Applying law of conservation to the system, you can conclude that kinetic energy is dissipated in the form of thermal energy. So,

$∆E_{th}=3.5\mathrm{KJ}$

Hence, there is an increase in the thermal energy of his body and the ground along which he slides.