Circular Work Vs. Straight line work

[kegger90]

I was having trouble understanding a concept in my physics class regarding work. If you have a 1000 lb weight being lifted 10 feet in the air would that have the same amount of work done as lifting a 1000 lb weight on a rigid pivot with a radius of 5 feet? if anyone could help explain this to me I would appreciate it. Thanks

 

[sophiecentaur]

The gain in Potential Energy is the same what ever (frictionless) path you take. If this were not so, you could make a machine that gave perpetual motion.

 

[LostConjugate]

The gain in Potential Energy is the same what ever (frictionless) path you take. If this were not so, you could make a machine that gave perpetual motion.

Work and energy are not equal though.

 

[sophiecentaur]

Same units, though (Joules). Potential energy is defined in terms of the work put in / got out. What do you mean by "not equal"?

 

[LostConjugate]

Same units, though (Joules). Potential energy is defined in terms of the work put in / got out. What do you mean by "not equal"?

Well in one case the object is going straight up while in the other it is following a longer path. I suppose the question is if this effects the work.

No machine can increase work (with the exception of my computer), so it should be the same, even though the force is multiplied by a longer distance, so the force must be less.

 

[jetwaterluffy]

Well in one case the object is going straight up while in the other it is following a longer path. I suppose the question is if this effects the work.

No machine can increase work (with the exception of my computer), so it should be the same, even though the force is multiplied by a longer distance, so the force must be less.

Yes, the force is less. Think about it. What would be easier, pushing a rock up a gentle slope, or lifting it straight up above your head?

 

[sophiecentaur]

Compare the work required to lift a weight vertically and up a slope - a simple case.
The work on the vertical lift is force times distance = mgh

On the slope, the force (down the slope) is mg sin(θ) where θ is the angle of the slope. The distance up the slope is h/sin(θ). So the force times the distance is mg sin(θ) h/sin(θ), which equals mgh again.
You can break any shape of uphill path into a series of small, straight sections. The above applies to each of the sections, showing that the work is independent of the path - just on the overall height gained. If that bit of maths wasn't too much of a pain, you can see it proves the point.