Question

a 54.0 - kg cheetah accelerates from rest to its top speed of 30.5 m/s. (a) how much net work (in j) is required for the cheetah to reach its top speed? (b) one food calorie equals 4186 j. how many calories of net work are required for the cheetah to reach its top speed? note: due to inefficiencies in converting chemical energy to mechanical energy, the amount calculated here is only a fraction of the energy that must be produced by the cheetahs body. details my notes ask your teacher practice another

Explanation:

Step1: Use work - energy theorem

The work - energy theorem states that the net work done on an object is equal to its change in kinetic energy. The initial velocity $v_0 = 0$ m/s and the final velocity $v = 30.5$ m/s, and the mass of the cheetah $m=54.0$ kg. The formula for kinetic energy is $K=\frac{1}{2}mv^{2}$.

Step2: Calculate the change in kinetic energy

$\Delta K=K - K_0$. Since $K_0 = 0$ (starts from rest), $K=\frac{1}{2}mv^{2}$. Substitute $m = 54.0$ kg and $v = 30.5$ m/s into the formula:
\[

$$\begin{align*} K&=\frac{1}{2}\times54.0\times(30.5)^{2}\\ &= 27\times930.25\\ &=25116.75\text{ J} \end{align*}$$

\]

Step3: Convert joules to calories

We know that $1$ calorie $ = 4186$ J. Let $x$ be the number of calories. Then $x=\frac{25116.75}{4186}\approx6.0$ calories

Answer:

(a) $25116.75$ J
(b) $6.0$ calories