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The following question was on a quiz in physics class:

If the net work done on a particle is zero, which of the following statements must be true?

a) The velocity is zero
b) The velocity is decreased
c) The velocity is unchanged
d) The speed is unchanged
e) There is no displacement for the object

The correct answer was e. In what scenario would the speed change?

There was an explanation beside the question:

Since the work done is zero, it indicates that the applied force is zero. Since Force = Mass X Acceleration, and the mass is not zero, this implies that the acceleration is zero.

When asked about this question, the teacher responded:

If there is no displacement, then only work done is zero. If the speed is unchanged, then there is no acceleration. This will lead to an absence of a force. Hope it explains the situation.

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  • $\begingroup$ Are you sure that is the correct answer? for I think it might be incorrect, and (d) would be more appropriate. $\endgroup$ – Satwik Pasani Sep 23 '13 at 4:24
  • $\begingroup$ Frankly, this is my thinking, but I always recognize the possibility that I'm completely missing something. The quiz was online and the grading is automated. I've even checked again before responding here; it definitely lists "There is no displacement for the object" as the correct answer. $\endgroup$ – roundar Sep 23 '13 at 4:28
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    $\begingroup$ What does the work-energy theorem tell you about net work? What do you know about the relationship of work and energy? I think the author of this question meant "if the net <b>nonconservative</b> work done on a particle is zero..." $\endgroup$ – Jerry Schirmer Sep 23 '13 at 4:30
  • $\begingroup$ (e) is incorrect unless there are other implicit assumptions, namely that there is a force acting on the particle and that the force is nonconservative. If there is no force acting, or if the force is conservative and the particle comes back to its starting point (or equivalent) then there is no work done. $\endgroup$ – Michael Brown Sep 23 '13 at 4:33
  • $\begingroup$ Not only is the teacher's answer wrong, the explanation is wrong too. It claims the applied force must be zero to have zero work. You can of course have a non-zero force perpendicular to the object's velocity, like a charged particle in a magnetic field. The magnetic force does no work, and the particle changes direction but not speed. $\endgroup$ – Tim Goodman Sep 23 '13 at 14:10
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If you've accurately described the problem and answer, your teacher made a mistake. The answer is d). Work is the change in kinetic energy, so if work is zero, kinetic energy is fixed and speed is fixed. Velocity can change by having the particle change direction. Velocity need not be zero, and therefore displacement need not be zero.

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Work is $\vec{F}.d\vec{s}$ so your teachers answer holds only if the force is nonzero. Otoh, if the force is zero, the object could have been displaced (for eg: moved at a constant speed). So unless there is some other information specifying that the external force is nonzero (for eg: multi part problem, or other information) the answer cannot be option (e).

Note that anyways, option (e) is a subset of option (d). If an object is not displaced, then its speed stays unchanged and is zero. So option (d) can not be wrong.

EDIT: The description in the picture seems crazy in multiple ways! O_o

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  • $\begingroup$ lol. Thank you. It gets worse, I had 10 minutes to do this quiz and the material is over a chapter we're doing weeks from now. $\endgroup$ – roundar Sep 23 '13 at 4:55
  • $\begingroup$ JerrySchirmer and MichaelBrown have good comments on the question. $\endgroup$ – Siva Sep 23 '13 at 5:00
  • $\begingroup$ Also note that option (a) is a subset of (e) and (d). If velocity is zero, displacement is zero and speed is unchanged. $\endgroup$ – tpg2114 Sep 23 '13 at 20:28
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In addition to the previous answers. You can have a good example showing that the answer (e) is incorrect. If you considered a charged particle entering a magnetic field. The magnetic field will NOT exert work on the particle, it will change its velocity but NOT its speed.

The particle will keep moving in the magnetic field although no work is exerted on it. A net displacement could be obtained if the magnetic field was chosen properly. That proves your teacher is wrong. The correct answer is (d)

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Using this from my book (Physics for Scientists and Engineers with Modern Physics 9th Ed):

If more than one force acts on a system and the system can be modeled as a particle, the total work done on the system is just the work done by the net force. If we express the net force in the x direction as o Fx, the total work, or net work, done as the particle moves from xi to xf is

ΣW = the integral of ΣF from xi to xf

A proof that d) the speed of the particle must be unchanged is true:

If ΣW = the integral of ΣF from xi to xf = 0, then Δx = 0 OR ΣF = 0 (this alone is proof that e) there must be no displacement is false). In the case that Δx = 0, the particle is not displaced and the speed is unchanged. If ΣF = ma0+ma1+ . . . + man = m(Σa) = 0 and m ≠ 0, then Σa = 0 and the speed is again unchanged. Therefore, if ΣW = 0, then it is necessarily true that the speed remains unchanged.

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  • $\begingroup$ I'll be accepting my own answer because, after I emailed my professor, he thanked me for the clarification and said he would change the quiz. If there is a problem with this answer, please point it out and I will change it and/or accept one of the most upvoted answers. $\endgroup$ – roundar Sep 24 '13 at 14:19
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Oh dear, I don't like it when physics teachers make conceptual mistakes like this. It certainly option d) the speed remains unchanged. Option e) is only valid if and only if the body is at rest with the inertial frame to begin with.

We cannot say velocity remains unchanged because the body could be rotating about an axis (change in direction alludes to change in velocity, but there is NO work done when an object is rotating about a fixed point).

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Does not matter if the speed change during the displacement. For example now you are on (x,y) location, if you go to the canada and come back your first location (x,y), your net work 0 whatever your speed with 500 km\s or 300 km\s

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