The forum discussion centers on the ambiguities present in high school physics problems, specifically regarding the calculation of distance using the equations for kinetic energy (E(k)=1/2mv^2) and work (W=fd). Participants highlight the lack of necessary information, such as mass and distance, to solve the problem accurately. The provided answer of 4,545.45 meters is debated, with contributors emphasizing the need for clearer problem statements that specify conditions like constant force and friction. The consensus is that without explicit details, assumptions must be made, which can lead to flawed interpretations.
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Since the velocity constant, the KE is constant, so that is not going to help.physicslady123 said:tried using the equation E(k)=1/2mv^2
Umm... the distance is what you are asked to find!physicslady123 said:I tried W=fd but there is no distance given
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 said: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?
Certainly that is true. But, given that there is an answer provided, one can assume that the simplifying assumptions hold.gneill said: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?
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 said:Certainly that is true. But, given that there is an answer provided, one can assume that the simplifying assumptions hold.
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.FactChecker said:Of course, one can imagine many complications that would invalidate that answer, but it appears to be the expected answer.
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 said: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.
Yes. I fully agree with that.haruspex said: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.
This is unnecessary. What is necessary is a specification of the 200 N being the position average rather than the time average.gneill said:Why should that be considered constant over time and speed?
FactChecker said: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.)
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 integralPeroK said:it doesn't really matter what the assumptions are.
I realize that, but the problem may not be a real-world problem.PeroK said: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.
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.That said, as long as you calculate ##W = Fd## it doesn't really matter what the assumptions are.
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##.FactChecker said:And the fact that the expected answer is the same as my simple answer confirms that that is what was desired.