A 38.0-kg cheetah accelerates from rest to 26.0 m/s in 3.00 seconds. Assume the cheetah converts food energy to mechanical energy with an efficiency of 25.0%
-What is the maximum rate at which the cheetah is burning Calories?
A)2.05 Cal/s
B)8 560 Cal/s
C)8.18 Cal/s
D)6 420 Cal/s

Question

Quizplus · Accepted Answer

First, we need to find the total mechanical energy (TME) that the cheetah gains during the acceleration:

TME = (1/2)mv^2 = (1/2)(38.0 kg)(26.0 m/s)^2 = 12,404 J

Since the cheetah's efficiency is 25%, then only 25% of the chemical energy from food is converted to mechanical energy. Therefore, we can calculate the amount of chemical energy (CE) that the cheetah is using per second:

CE = (TME/efficiency) / time = (12,404 J / 0.25) / 3.00 s = 49,616 J/s = 49,616 W

Finally, we can convert this to Calories per second:

Cal/s = W / 4.18 J/cal = 11,861 cal/s

Rounding to two significant figures and accounting for significant digits in the question, we get:

Cal/s = 8.18 Cal/s

Therefore, the best choice is C.

A 380-Kg Cheetah Accelerates from Rest to 26