Is Constant Velocity Assumed in Work-Energy Problems?

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SUMMARY

The discussion clarifies that in work-energy problems, one cannot assume constant velocity without explicit information indicating zero net force. The relationship established is that zero net force correlates with constant kinetic energy, while a non-zero net force indicates a change in kinetic energy. The forces acting on the box moving down the incline—weight, normal force, and friction—do not provide sufficient information to conclude constant velocity. Therefore, assumptions about net force and kinetic energy must be based on the specific details provided in the problem.

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  • Understanding of Newton's laws of motion
  • Familiarity with work-energy principles
  • Knowledge of forces acting on objects (weight, normal force, friction)
  • Basic concepts of kinetic energy and net work
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mathsciguy
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It's been a while since I've done any work-energy questions, and I just noticed how vague some questions could be (or maybe I don't understand it that well). Anyway, suppose I have a box moving down an incline where the only forces acting on it are its weight, the normal force, and friction. Now, can I assume that this box is moving in constant velocity, or there is zero net force? Or is this something that is apparent and it does not need to be assumed since in this particular question the assumption was never given, but the known variables seem fit the case where there is constant velocity.

Thanks.
 
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mathsciguy said:
... work-energy questions, ... suppose I have a box moving down an incline where the only forces acting on it are its weight, the normal force, and friction. Now, can I assume that this box is moving in constant velocity, or there is zero net force?
There is no reason from your description to assume a constant velocity. And, from your freshman physics, constant velocity ⇔ zero net force, so there is neither reason to assume zero net force.

net force = 0 ⇔ net work = 0 ⇔ kinetic energy remains constant.

net force ≠ 0 ⇔ net work ≠ 0 ⇔ kinetic energy changes.

Which happens, and increase or decrease in kinetic energy, depends on more information that is not given in your description.
 
Oh yeah, that was really silly. Thanks.
 

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