Physics Professor Bicycles Through Air: Power Calculations

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The discussion revolves around a physics problem involving a professor cycling through air, focusing on calculating the mass of air swept out and the power required to accelerate it. The correct mass of air swept out in one second is 6.45 kg, confirmed through proper calculations. However, the initial attempt to find the power required was flawed, leading to confusion over the correct value, which is actually 323 watts, derived from understanding average power based on air acceleration. A follow-up question reveals that, considering the professor's aerodynamic shape, the power required to accelerate the air is reduced to 52.4 watts, and with a 20% efficiency, he would burn 676 kilocalories in three hours. The key takeaway is the importance of correctly applying physics equations and understanding average power in these calculations.
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Homework Statement



A physics professor bicycles through air at a speed of v= 36 km/hr. The density of Air is 1.29 kg per m^3. the professor has a cross section of 0.5 m^2. assume all of the air the professor swept out is accelerated to v.

a.) What is the mass of the air swept out by the professor in one second? (ans. 6.45 kg)

b.) What is the power to required to accelerate this air? (323 w)

Homework Equations



p=w/t
P= Fv

The Attempt at a Solution



Multiplying the cross section to the velocity 10 m/s(36km/hr) divided by 1 sec , i got 5 m^3.. and then multiplying it to the denisty, i got 6.45 kg.. the correct ans was 6.45 kg... but somehow i got a feeling my solution is flawed... is my solution is correct and also i can't find the correct answer in letter b.. I tried substituting P= Fd/t or P=ma/t and gor 64.5 W, which is wrong... can someone tell me where did i go wrong in finding the power required to accelerate in air...?
 
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scytherz said:

Homework Statement



A physics professor bicycles through air at a speed of v= 36 km/hr. The density of Air is 1.29 kg per m^3. the professor has a cross section of 0.5 m^2. assume all of the air the professor swept out is accelerated to v.

a.) What is the mass of the air swept out by the professor in one second? (ans. 6.45 kg)

b.) What is the power to required to accelerate this air? (323 w)

Homework Equations



p=w/t
P= Fv

The Attempt at a Solution



Multiplying the cross section to the velocity 10 m/s(36km/hr) divided by 1 sec , i got 5 m^3.. and then multiplying it to the denisty, i got 6.45 kg.. the correct ans was 6.45 kg... but somehow i got a feeling my solution is flawed... is my solution is correct
your solution is correct, but you didn't divide by 1 sec, you multiplied by 1 second to get the length of the air column
and also i can't find the correct answer in letter b.. I tried substituting P= Fd/t or P=ma/t and gor 64.5 W, which is wrong... can someone tell me where did i go wrong in finding the power required to accelerate in air...?
Doing your math, you should have got 645 watts, which is twice the answer...You are looking for average power, which is based on average speed of the air as it accelerates from 0 to 10 m/s.
 
PhanthomJay said:
your solution is correct, but you didn't divide by 1 sec, you multiplied by 1 second to get the length of the air column Doing your math, you should have got 645 watts, which is twice the answer...You are looking for average power, which is based on average speed of the air as it accelerates from 0 to 10 m/s.

Yeah your right! thank you! By the way there was a follow up question on that...

Homework Statement



If the power required to accelerate the air is 40% of the answer from the last problem due to the professor's sleek aerodynamic shape,

a) what is the power required to accelerate the air? ans 52.4 W



b) If the professor has an efficiency of 20%, how many kilocalories will he burn in three hours?
DATA: 1 kcal=4187 J

ans 676 kcal

Homework Equations



P=W/t
P= Fv


The Attempt at a Solution



As you can see, the power requires in the previous question was 323 watts, so I just multiply it on .40 (40%) and got 129.2 w, which is wrong... the real answer is 52.4 w... I even tried multiplying the density of air to .40 but still got a wrong answer...
 

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