A 59.0 kg cheetah accelerates from rest to its top speed of 32.7 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 cheetah's body.

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Asked Dec 10, 2019
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A 59.0 kg cheetah accelerates from rest to its top speed of 32.7 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 cheetah's body.
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Expert Answer

Step 1

a)

Work done is equal to change in kinetic energy.

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w = (0.5)(59.0kg)(32.7 m/s) -(0m/s)*) = 3.15 x10ʻ J

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Step 2

Answer: 3.15×104 J

Step 3

b)

Work done in calo...

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1 food calorie W = 3.15 x 10 J 4186J = 7.52 food calorie

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