Power Homework: Mass & Power Calculation Problem

[scytherz]

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= 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...?

 

[anathema]

Your solution for part a) is correct. The mass of the air swept out by the professor in one second is indeed 6.45 kg.

For part b), you can use the formula P = Fv to find the power required to accelerate the air. In this case, the force (F) would be the mass of the air (6.45 kg) multiplied by the acceleration (v), which is the same as the velocity of the professor (36 km/hr or 10 m/s). So the power required would be 6.45 kg x 10 m/s = 64.5 W.

However, this is the power required to accelerate the air. The question is asking for the power required to accelerate all the air swept out by the professor, which would be the power required multiplied by the number of air molecules swept out per second. This can be calculated by dividing the mass of the air (6.45 kg) by the mass of one air molecule (1.29 x 10^-26 kg), giving a total of 4.98 x 10^25 air molecules swept out per second. Multiplying this by the power required to accelerate one air molecule (64.5 W), we get a total power requirement of 3.23 x 10^27 W, or 323 W as given in the answer.

Hope this helps clarify your solution. Keep up the good work in your studies of physics!