Calculating Work Done on a Slope: Friction and KE

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When pushing a box up a slope at constant speed, the work done by friction cannot be assumed to be zero, despite the constant velocity. The frictional force is typically dependent on the weight of the box and the coefficient of friction, not on acceleration. Therefore, the work done by friction is calculated as the frictional force component parallel to the surface multiplied by the distance traveled. The work done by non-conservative forces can be equated to the change in gravitational potential energy, with no change in kinetic energy. Understanding these dynamics is crucial for accurate calculations in physics.
disruptors
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Hey ppl,

If you have a man pushing a box up a slope with a horizontal force at constant speed from distance a to b, can one assume the Work done by friction to be zero all the time since velocity is constant? or can one assume W(neoconservative)=change in U(gravitational PE) only, with change in KE=0...

Thanks
 
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disruptors said:
can one assume W(neoconservative)=change in U(gravitational PE) only, with change in KE=0...

Yes, but only if the man is a neoconservative. :-p
 
hahaha :smile:

disruptors said:
Hey ppl,

If you have a man pushing a box up a slope with a horizontal force at constant speed from distance a to b, can one assume the Work done by friction to be zero all the time since velocity is constant?

Well that would be true if the frictionnal force depended on the accelaration of the thing sliding on the surface. But it's usually not the case, and we can assume that the frictionnal force depends only on the weight of the box and on the coeficient of friction of the surface, such that f = Nµ = mgµ. So the work done by the fricton would have to be component of friction parallel to the surface times the distance b-a.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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