Clarifying Ambiguities in High School Physics Problems

[physicslady123]

Homework Statement

An average force of 220N is required to keep a motorcycle moving in uniform velocity. How far will one MJ of energy take the motorcycle?

Relevant Equations

P=ΔE/t or P=W/t or P=Fd/t

I tried using the equation E(k)=1/2mv^2 and isolating for v but no mass was given. Then, I tried W=fd but there is no distance given. I don't know how to solve this.

 

[gneill]

Yep. Without knowing the velocity that the 220 N sustains, it's not possible to arrive at a calculated answer.

Is your problem description a complete and accurate statement, word for word, of the problem as it was given to you?

 

[physicslady123]

Yep. Word for word. My teacher posted the answer: 4.5*10^3m is the distance

 

[gneill]

Well, I don't see any way forward with the problem as given; there's not enough information in the problem statement to proceed. Perhaps you can inquire about the problem to your teacher (or TA, if there is one)?

 

[haruspex]

tried using the equation E(k)=1/2mv^2

Since the velocity constant, the KE is constant, so that is not going to help.

@gneill is right that there is not enough information, because it specifies average force. If you assume the force is constant there is a solution.

I tried W=fd but there is no distance given

Umm... the distance is what you are asked to find!

 

[FactChecker]

I think I interpret the problem differently:
It takes 220 N force to keep it moving at a constant velocity. (I guess that is being opposed by 220 N of combined friction, drag, gravity, etc.) If the motorcycle has a certain amount of (kinetic) energy, how far will it go if the 220 N force that keeps it moving is removed? One must assume that the force opposing the motion does not change.

PS. The provided answer is nearly correct for this problem (4545.45 meters is what I get.)

 

[gneill]

I think I interpret the problem differently:
It takes 220 N force to keep it moving at a constant velocity. (I guess that is being opposed by 220 N of combined friction, drag, gravity, etc.) If the motorcycle has a certain amount of (kinetic) energy, how far will it go if the 220 N force that keeps it moving is removed?

Surely that would depend upon its initial velocity, friction (of all kinds) , the path of the vehicle (is it traveling along straight horizontal path or one that is rising, falling or even oscillating in elevation). None of these items are detailed in the problem statement. What "certain amount of KE " can you ascribe to the undefined motorcycle?

 

[FactChecker]

Surely that would depend upon its initial velocity, friction (of all kinds) , the path of the vehicle (is it traveling along straight horizontal path or one that is rising, falling or even oscillating in elevation). None of these items are detailed in the problem statement. What "certain amount of KE " can you ascribe to the undefined motorcycle?

Certainly that is true. But, given that there is an answer provided, one can assume that the simplifying assumptions hold.

 

[gneill]

Certainly that is true. But, given that there is an answer provided, one can assume that the simplifying assumptions hold.

I'm not clear on what any simplifying assumptions there may be here. As far as I can see, there is insufficient given information to proceed. I'd be very happy to be shown otherwise!

 

[FactChecker]

Force opposing motion = -Force to keep it going at a constant speed = -220 N
Remove the force to keep it going, leaving only the force opposing motion.
1 MJ = Work = Force opposing motion * distance = 220 N * distance.
distance = 1,000,000/220 = 4,545.45 meters

Of course, one can imagine many complications that would invalidate that answer, but it appears to be the expected answer.

 

[gneill]

Of course, one can imagine many complications that would invalidate that answer, but it appears to be the expected answer.

Agreed. But the question itself is seriously flawed in that it does not explicate the vehicle's path nor the relationship between the vehicle's velocity and any frictional resistance to its motion. Why should that be considered constant over time and speed? There are no obvious reasons presented.

 

[haruspex]

Agreed. But the question itself is seriously flawed in that it does not explicate the vehicle's path nor the relationship between the vehicle's velocity and any frictional resistance to its motion. Why should that be considered constant over time and speed? There are no obvious reasons presented.

It would have been better if it had asked how far does each 1MJ take the bike, and specified a constant retarding force. Nothing more would have been needed.

 

[gneill]

It would have been better if it had asked how far does each 1MJ take the bike, and specified a constant retarding force. Nothing more would have been needed.

Yes. I fully agree with that.

 

[Orodruin]

Why should that be considered constant over time and speed?

This is unnecessary. What is necessary is a specification of the 200 N being the position average rather than the time average.

 

[PeroK]

I think I interpret the problem differently:
It takes 220 N force to keep it moving at a constant velocity. (I guess that is being opposed by 220 N of combined friction, drag, gravity, etc.) If the motorcycle has a certain amount of (kinetic) energy, how far will it go if the 220 N force that keeps it moving is removed? One must assume that the force opposing the motion does not change.

PS. The provided answer is nearly correct for this problem (4545.45 meters is what I get.)

Drag certainly isn't independent of speed, so I don't see how you can assume that the drag force remains constant as the bike slows to rest.

That said, as long as you calculate $W = Fd$ it doesn't really matter what the assumptions are.

 

[Orodruin]

it doesn't really matter what the assumptions are.

It does matter if it is a time average rather than a distance average. For a time average $\bar F$you cannot do $W = \bar Fd$. To compute work you would have to do the integral
$$
\int F \, dx = \int F v\, dt.
$$
Since $v$ is a function that depends on $F$, this is not going to be proportional to $\bar F$.
For a position average $\bar F$, you have by definition that
$$
\bar F = \frac 1d \int F\, dx = \frac Wd
$$
so in that case $W = \bar F d$ works fine.

 

[FactChecker]

Drag certainly isn't independent of speed, so I don't see how you can assume that the drag force remains constant as the bike slows to rest.

I realize that, but the problem may not be a real-world problem.

That said, as long as you calculate $W = Fd$ it doesn't really matter what the assumptions are.

Of course. But unless you specify the value of F through the entire time, the answer can not be determined. I have assumed a constant force and ignored brakes, drag, variations in terrain, variations in throttle, etc., etc., etc. I think that is reasonable in an elementary physics question. And the fact that the expected answer is the same as my simple answer confirms that that is what was desired.

 

[Orodruin]

And the fact that the expected answer is the same as my simple answer confirms that that is what was desired.

I have found historically that high school teachers are often not the best in being precise in formulating problems in a way such that people with more advanced knowledge than their students do not find ambiguities. The underlying unstated assumptions in the questions are often on the level such that the problem becomes solvable by a high school student, which in this case means it is safe to assume $W = Fd$.