How Does Force x Distance Relate to Potential Energy and Work Done?

[primarygun]

Work Done Sorry, Still have some Qs

Work done refers to a force x distance.
Or a force (continuous) x distance?
I think it is the second one when I think of the potential energy, right?
For PE=mgh,
How to prove it? And when an object falls, it loses potential energy, why?
Where is the negative sign?

 

[primarygun]

How to determine F=Weight? Isn't that if you apply mg upward, the distance will be 0?
So we can only think of it is falling with a force mg and the distance traveled but not think of it being elevated for the proof of the equation?

 

[Mellow^Guy]

you know that the work done by a conservative force does not depend on path

this is how to prove it:

 

[primarygun]

Potential energy has no sign since it is not a vector.
Work= constant force (continuous)x distance.
If it is added by a continuous force, the object should not stopped no matter there is friction(smaller than applied force). So not continuous force?

 

[primarygun]

In $$ KE=1/2mv^2 $$
Why the proof is using u=0 but not v=0? Isn't after the work is done, v=0?

 

[James R]

Work done by a constant force \vec{F}, acting over a displacement \vec{s} is:

W = \vec{F}.\vec{s}

Take a constant gravitational force acting on a mass: \vec{F} = m\vec{g}. The force acts to make the mass fall through a displacement of \vec{s} = \vec{h}, where \vec{h} is directed downwards. The work done by gravity on the mass as it falls is:

W = m\vec{g}.\vec{h} = mgh,

which is positive since the force and the displacement vectors are in the same direction.

Gravity is a conservative force, and the change in potential energy associated with such a force is the negative of the work done by the force. i.e.

\Delta U = -W

For gravity

\Delta U = -mgh,

so the potential energy decreases as the mass falls.

 

[HallsofIvy]

What are you talking about?

It would help if, when you ask "Why the proof is using u=0 but not v=0? Isn't after the work is done, v=0?", you tell us what u and v are. I assume that v is the speed of some object. In that case, yes, the kinetic energy of an object of mass m is \frac{1}{2}mv^2. I have no idea what "u" is. IF work is done to bring the object to a halt, then, yes, v= 0 when it is no longer moving. I don't know why whatever problem you are talking about doesn't take v= 0 because I don't know what problem you are talking about.

"If it is added by a continuous force, the object should not stopped no matter there is friction(smaller than applied force). So not continuous force?"

I assume you are talking about a problem in which there is a friction force, f_e, and an "applied force", f_a, opposite to the friction. The net force will be f_a-f_e. Assuming that f_a> f_e, the not come to a halt. In fact it will keep accelerating. How about posting a specific problem so we can see what you are talking about?

 

[primarygun]

I think the force in the equation of work done is continuously applied,like weight, right?
And the object would not stop finally, right?

For the PE, a weight of 100N is lifted up by a man up to a height of 10m.
Why the force applied by the man continuously is 100N?
Isn't 100 N causing no effect on the weight as it is compensated by the mg of the weight, right?