Question

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?

Short answer

1. Decrease in the player’s kinetic energy is 3.5 KJ .
2. Increase in the thermal energy of his body and the ground along which he slides is $∆E_{th}=3.5\mathrm{KJ}$.

Step-by-step solution

Given data:

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

To understand the concept:

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) Calculate the decrease in the player’s kinetic energy:

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 .

(b) Calculate the increase in the thermal energy of his body and the ground along which he slides:

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.