Work and Portential Energy (U)

[clipperdude21]

Just a quick question. If you are pushing an object to the right with a force on a surface with friction and the object is moving at constant velocity to the right... the work of the force is positive so does that mean delta U is positive as well since Work of conservative forces= delta U?

Thanks!

 

[PhanthomJay]

Theer are no conservative forces acting. (BTW, the work done by conservative forces, when they exist, is - delta U).

 

[clipperdude21]

9. A conservative force F is directed along the horizontal x direction with F(x) = (2.0x
+4.0)N (where x is expressed in meters). It acts on a 5 kg mass that moves at constant
velocity from x=1.0 m to x=5.0 m. The coefficient of kinetic friction of the mass with the
surface is 0.1.

that was the homework problem. what i did was integrate F and made it negative since -du/dr=fr(r) so U(r) is -X^2 - 4x.

The last part asks about what is the change in potential energy of the force. I just took U(1)-U(5) which i think is right from what you said... W= -delta U.

 

[PhanthomJay]

9. A conservative force F is directed along the horizontal x direction with F(x) = (2.0x
+4.0)N (where x is expressed in meters). It acts on a 5 kg mass that moves at constant
velocity from x=1.0 m to x=5.0 m. The coefficient of kinetic friction of the mass with the
surface is 0.1.

that was the homework problem. what i did was integrate F and made it negative since -du/dr=fr(r) so U(r) is -X^2 - 4x.

The last part asks about what is the change in potential energy of the force. I just took U(1)-U(5) which i think is right from what you said... W= -delta U.

Oh, ok, when you said the mass was being pushed to the right, I was envisioning an applied non-conservative force, sorry. But just be certain of your plus and minus signs. Since the work done by the conservative force is positive (i.e., force and displacement are in same direction), then the PE change of that force must be negative. It's like dropping a mass off a building from height h, where the work done by the conservative gravity force is +mgh, and the PE change is -mgh.

 

[clipperdude21]

thanks! i get how when an object is dropped its PE change is negative but I am not sure why if you push an object horizontally to the right with a force, the PE change would be negative

 

[clipperdude21]

oh and one more question just to make it clear for me:

W of conservative forces = - delta(PE) right? since that would make my answers make sense.

 

[clipperdude21]

is W= -delta(PE) an equation?

 

[PhanthomJay]

thanks! i get how when an object is dropped its PE change is negative but I am not sure why if you push an object horizontally to the right with a force, the PE change would be negative

The work done by conservative forces is the negative of the PE change. This follows from the conservation of energy and work energy theorems. Since
$$W_{nc} = \Delta{KE} + \Delta{PE}$$ (conservation of energy) and
$$W_{total} = W_{nc} + W_{c} = \Delta{KE}$$ (work-energy theorem), then
$$W_{c} = -\Delta{PE}$$
To answer your question, let's say the conservative force pushing the object horizonatlly is a compressed spring that is released. Initially it has a PE of 1/2kx^2. When released, it does positive work on the object equal to 1/2kx^2 when the object loses contact with the spring, at which point the spring is now back to its original uncompressed length and has no potential energy left, that is,
the change in PE is 0 - 1/2kx^2 = -1/2kx^2.
Does this make sense to you?