Conservation of energy: Conceptual question

[Neha Siddhartha]

Homework Statement

Two identical twins, A & B, are riding identical bikes up the same hill, both at constant speed. Twin A takes 20 seconds to climb the hill, while twin B takes 40 seconds.

a) Neglecting all forms of friction, which twin consumes more energy?

b) With friction, which twin consumes more energy?

Homework Equations

Kinetic E = ½mv[2]
Gravitational Potential E = mgh

The Attempt at a Solution

a) Intuitively, I know that both consume the same amount of energy.
Their gain in potential energy is the same, as they are going up the same hill so h, or height, must be the sa
me.
However, I do not understand how to explain it using the equations. Twin A would have 4x the kinetic energy of Twin B, so the kinetic E applied does not equal the potential E gained. (½mv[2] ≠ mgh).
So, should I instead think of the total energy of each biker?:
Total E consumed = ½mv[2] + mgh

b) The biker going faster (Twin A) would consume more energy. I think that with a greater velocity they will encounter greater friction. (Is this correct?) My professor said friction is just random or nondirectional kinetic energy. How can I explain this mathematically with just the equations given?

 

[berkeman]

a) Neglecting all forms of friction, which twin consumes more energy?

b) With friction, which twin consumes more energy?

Welcome to the PF.

I don't think you mean to use the word "friction", since it's impossible to "climb a hill" without friction with the ground. I think you mean "air resistance", right? Then the answer to part (b) is more straightforward.

 

[rude man]

Hint: twin "B" must be taking the long way around, as he travels twice as far as twin "A". What's that suggest as far as friction is concerned (not air but tire-to-ground)?

 

[jbriggs444]

Hint: twin "B" must be taking the long way around, as he travels twice as far as twin "A". What's that suggest as far as friction is concerned (not air but tire-to-ground)?

I disagree with this hint.

The phrase "constant speed" indicates that each twin stays at the same speed she started with. It does not indicate that both twins travel at the same speed as each other. Accordingly, it is entirely possible that both twins are following the same route up the hill. That is the assumption I make when reading the problem -- that the route is the same and one twin is moving twice as fast as the other. [That said, the standard rule of test taking applies -- if you have to make assumptions, state them clearly up front and proceed with the rest of the problem. That way you are likely to get credit even if your interpretation of the problem disagrees with the instructor's]

 

[CWatters]

Neha - Think about the work done in each case. For example what is the equation for the work done against gravity? Friction? I think you might be over complicating the answer.

Note it only mentions friction not air resistance/drag.

 

[rude man]

I disagree with this hint.

The phrase "constant speed" indicates that each twin stays at the same speed she started with. It does not indicate that both twins travel at the same speed as each other. Accordingly, it is entirely possible that both twins are following the same route up the hill. That is the assumption I make when reading the problem -- that the route is the same and one twin is moving twice as fast as the other. [That said, the standard rule of test taking applies -- if you have to make assumptions, state them clearly up front and proceed with the rest of the problem. That way you are likely to get credit even if your interpretation of the problem disagrees with the instructor's]

On second thoughts I agree with you. Thanks.

 

[Neha Siddhartha]

Neha - Think about the work done in each case. For example what is the equation for the work done against gravity? Friction? I think you might be over complicating the answer.

Note it only mentions friction not air resistance/drag.

My professor said that changes in E do work. In (a) the velocity and therefore the kinetic E of both twins stay the same, so you only look at the gain in gravitational potential energy (E = mgh). Therefore, to balance this identical gain in E, they must consume the same amount of E.

(b) represents the real world version, where we lose E to friction.
My professor defines air friction as heat, which is random kinetic energy. So does a greater directed kinetic E translate to greater amount of heat?
Also, what is the equation that I should be using to define friction?

Thank you! :)

 

[rude man]

For part b, friction between the tire and ground will be what - the same or different? (Hint: use coefficient of kinetic friction, generally assumed independent of speed).
Then there is the other friction - air movement - see post 2!

 

[rude man]

Also, what is the equation that I should be using to define friction?
Thank you! :)

Air friction force: F = kv² typically assumed, v = velocity.

 

[CWatters]

Ask your teacher if you should include air resistance in pary b or not.

Friction is normally considered independent of speed. Air resistance does depend on speed.