Question

A 75-kg sprinter accelerates from rest to a speed of 11.0 m/s in 5.0 s. (a) calculate the mechanical work done by the sprinter during this time. (b) calculate the average power the sprinter must generate. (c) if the sprinter converts food energy to mechanical energy with an efficiency of 25%, at what average rate is he burning calories? (d) what happens to the other 75% of the food energy being used?

Answer 1

The mechanical work done by the sprinter during this time will be 4537.5 J , the average power the sprinter must generate will be 907.5 W and if the sprinter converts food energy to mechanical energy with an efficiency of 25% then he will be burning calories at 54.20 calories per second.

Work in physics is the energy that is transferred to or from an item when a force is applied along a displacement. It is frequently described in its most basic form as the result of force and displacement.

The quantity of energy moved or transformed per unit of time is known as power in physics. The watt, or one joule per second, is the unit of power in the International System of Units.. A scalar quantity is power.

Given 75-kg sprinter accelerates from rest to a speed of 11.0 m/s in 5.0 s.

So let,

m = 75 kg

v = 11.0 m/s

t = 5.0 s

So the mechanical work done by the sprinter during this time will be as follow:

W = 0.5 mv²

W = 0.5 (75)(11)²

W = 4537.5 J

The average power the sprinter must generate will be as follow:

Power(P) = W / t

P =  4537.5/5

P = 907.5 W

Only 25% is absorbed. So, the sprinter only absorbed 226.875 J per second which is equal to 54.20 calories per second.

Hence   mechanical work done by the sprinter during this time will be 4537.5 J , the average power the sprinter must generate will be 907.5 W and if the sprinter converts food energy to mechanical energy with an efficiency of 25% then he will be burning calories at 54.20 calories per second.

Learn more about mechanical power here:

brainly.com/question/25573309

#SPJ10