What Is the Total Work Done on a Cyclist Ascending an Incline?

[student34]

Homework Statement

A man on his bike have an initial velocity of 8.5m/s. He begins to go up an incline that has a vertical height of 5.2m. Once he reaches the top of the incline, he only has a velocity of 3.00m/s. The man and his bike combined weigh 80kg. What is the total work done on the man from the bottom of the incline to the top?

Homework Equations

Ek1 + U1 + W = Ek2 + U2

The Attempt at a Solution

0.5(80kg)(8.5m/s)^2 + 0 + W = 0.5(80kg)(3m/s)^2 + 5.2m(9.81m/s^2)(80kg)

W = 0.5(80kg)(3m/s)^2 + 5.2m(9.81m/s^2)(80kg) - 0.5(80kg)(8.5m/s)^2

W = 1550.96J

But that is definitely the wrong answer. I can't think of any way that to get the right answer.

 

[Matejxx1]

are we suppose to just ignore friction?
because when the man is traveling uphill there will be friction acting against the way he is moving

 

[Orodruin]

Why do you think the answer is wrong?

Technically, it is impossible to know the answer here because the question only specifies the mass of man+bike and wants to know the work done on the man. Edit: Of course, the intent of the question is probably to ask what work is being done on man+bike ...

You should also round your answer to a reasonable number of significant digits. Using six significant digits is far too much unless you know all parameters to a precision of six digits (for example, the mass of the man to an accuracy of a few mg).

are we suppose to just ignore friction?
because when the man is traveling uphill there will be friction acting against the way he is moving

Friction is irrelevant. You know the initial and final energies and the work done on the man includes any possible negative work done by friction.

 

[student34]

are we suppose to just ignore friction?
because when the man is traveling uphill there will be friction acting against the way he is moving

Yeah, the question said to ignore friction.

 

[student34]

Why do you think the answer is wrong?

Technically, it is impossible to know the answer here because the question only specifies the mass of man+bike and wants to know the work done on the man. Edit: Of course, the intent of the question is probably to ask what work is being done on man+bike ...

Yes, sorry, it asks what the work done on the man and the bike is.

 

[Orodruin]

Yes, sorry, it asks what the work done on the man and the bike is.

This still does not answer the relevant question in my post: Why do you think your answer is wrong?

 

[student34]

This still does not answer the relevant question in my post: Why do you think your answer is wrong?

My physics professor uses an online assignment marker. It marked my answer wrong. It gives hints if the answer is close or just off by some sig-digs; it did not give me any hints.

 

[Matejxx1]

Does he start at h=0?

 

[Matejxx1]

Does he start at h=0?

Well that doesn't actually matter

 

[haruspex]

This still does not answer the relevant question in my post: Why do you think your answer is wrong?

I've always been a bit confused about this, but is it not more correct to say that the work done in getting up the hill was done on the Earth-man-bike system, rather than on the man&bike system? To put it another way, something (engine? muscles?) did 1551J of work, but gravity also did negative work on man&bike, so the net work done on them is just $\Delta KE$.

 

[Orodruin]

I've always been a bit confused about this, but is it not more correct to say that the work done in getting up the hill was done on the Earth-man-bike system, rather than on the man&bike system? To put it another way, something (engine? muscles?) did 1551J of work, but gravity also did negative work on man&bike, so the net work done on them is just $\Delta KE$.

I would say this depends on where you consider gravitational energy is "stored". In the case of considering man+bike in a setting where the gravitational field is a background, it makes some sense to assign the potential energy to the man+bike system. In other cases it would depend on how you treat the potential.

You might be correct in that this might be the intention of the questioner, but this needs to be specified.

 

[haruspex]

I would say this depends on where you consider gravitational energy is "stored". In the case of considering man+bike in a setting where the gravitational field is a background, it makes some sense to assign the potential energy to the man+bike system. In other cases it would depend on how you treat the potential.

You might be correct in that this might be the intention of the questioner, but this needs to be specified.

A quick search on Work-Energy Theorem suggests work done on a body is only the KE change. See e.g. http://faculty.wwu.edu/vawter/physicsnet/topics/Work/WorkEngergyTheorem.html

 

[Dodsy]

The answer should just be Eg, given the law of conservation of energy.Et = Eg + EkEt = EgEg= mghEg = (80kg)(9.8n/kg)(5.2m)

Eg = 4'076.8 J

Just a thought, if I'm wrong so be it.To prove this you can see how fast he would be going if he went down the hill. Take whichever number you get and add it to this formula :Etotal = W = EkEk = ½ mv²W = ½ (80kg)v²V²= (2(4076.8))/(80.0kg)V² = 101.92V= 10.09 m/sIf I am wrong, I apologize.

 

[Orodruin]

A quick search on Work-Energy Theorem suggests work done on a body is only the KE change. See e.g. http://faculty.wwu.edu/vawter/physicsnet/topics/Work/WorkEngergyTheorem.html

Yes, you are right. Taking the potential energy separate from the man+bike system does make sense. I am swayed to the opinion of adding the negative work done by gravity to end up with the difference in kinetic energies.

 

[Orodruin]

The answer should just be Eg, given the law of conservation of energy.

No, this is not how work done works. What you are computing is just the work related to the gravitational force.

 

[Orodruin]

The answer should just be Eg, given the law of conservation of energy.

No, this is not how work done works. What you are computing is just the work related to the gravitational force.