# Physics Professor Bicycles Through Air: Power Calculations

• scytherz
In summary, a physics professor bicycles through air at a speed of 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. Assuming 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? b) what is the power to required to accelerate this air?
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)

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...?

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)

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

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...

## 1. How is power calculated in physics?

In physics, power is defined as the rate at which work is done or energy is transferred. It is calculated by dividing the amount of work done by the time it takes to do it. The unit of power is watts (W), which is equal to one joule per second.

## 2. What is the role of a physics professor in calculating power?

A physics professor is an expert in the field of physics and is knowledgeable about the principles and equations used to calculate power. They are responsible for teaching students how to apply these equations to real-world scenarios, such as the power calculations involved in bicycling through air.

## 3. How does air resistance affect power calculations in cycling?

Air resistance, also known as drag, is a force that opposes the motion of an object through air. When bicycling through air, the rider must overcome this resistance, which requires more power. The amount of power needed to overcome air resistance depends on factors such as the speed of the bike, the surface area of the rider, and the density of the air.

## 4. How does the power needed to cycle through air change with different biking positions?

The power needed to cycle through air can vary depending on the position of the rider. For example, a rider in a more upright position will have a larger surface area facing the air, which will result in more air resistance and require more power. On the other hand, a rider in a more aerodynamic position will have a smaller surface area and less air resistance, requiring less power.

## 5. What other factors besides air resistance affect power calculations in cycling?

Besides air resistance, other factors that can affect power calculations in cycling include the weight and speed of the rider, the terrain, and the type of bicycle being used. These factors can all impact the amount of work that needs to be done and therefore the power required to overcome it.

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