work+conservation of energy

[UrbanXrisis]

Sliding a block on a floor that has friction. Going from points A to B to C to D then back to A. points ABCD make a rectangle. Moving the block at a constant speed. What is the work done by the closed loop?

Since speed is a constant, then there isn't a change in KE, so then work done has to be zero right?

What if I did this on a frictionless table? Would the work done be zero as well?

[Doc Al]

Sliding a block on a floor that has friction. Going from points A to B to C to D then back to A. points ABCD make a rectangle. Moving the block at a constant speed. What is the work done by the closed loop?
The work done by what force? To move that block a force must be applied to overcome friction. That force does work which ends up increasing the internal energy of the block + floor. (They heat up.)

Since speed is a constant, then there isn't a change in KE, so then work done has to be zero right?
The work-energy theorem says that the work done by the net force on an object will equal the change in KE. But the net force is zero, thus no change in KE.

What if I did this on a frictionless table? Would the work done be zero as well?
If you applied the same force, the block would accelerate and the work done would equal the change in KE of the block. To move the block at a constant speed requires no force (ignoring those corners!).

[arildno]

"Since speed is a constant, then there isn't a change in KE, so then work done has to be zero right?"

The work of the NET force is zero, but this doesn't mean that the work from the individual forces (friction and pushing force) are each zero.
The individual forces' work is most certainly not zero, only their SUM.

[disruptors]

"Since speed is a constant, then there isn't a change in KE, so then work done has to be zero right?"

Work isn't zero for individual forces just like urban said. On a frictional surface conservative energy isn't conserved thus W(neoconservative)=change(mgh)+0 KE...The force is neoconservative in this case I think thus W(nc of friction)=mgh if you have those variables of h and m.

[UrbanXrisis]

If you applied the same force, the block would accelerate and the work done would equal the change in KE of the block. To move the block at a constant speed requires no force (ignoring those corners!).

Does this mean there is no change in KE?

[Doc Al]

Does this mean there is no change in KE?
If the net force is zero, no work is done and the KE does not change.

[UrbanXrisis]

so there is no change in KE for frictionless or friction surfaces?

How do you know if energy is conserved or not?

[Doc Al]

so there is no change in KE for frictionless or friction surfaces?
If the object moves at constant speed how can the KE change? :smile:

How do you know if energy is conserved or not?
I'm not sure what you mean:
(1) For the frictionless surface, no energy is added (no work is done by an outside force) so energy is certainly conserved.
(2) For the friction surface, work must be done by an applied force. That work goes to heat. But no net work is done on the object, so its mechanical energy is conserved.

[gschjetne]

Hey, couldn't you just use this formula:

W = ( \mu mg \cdot \vec{F}_\textrm{normal} ) ( AB + BC + CD + DA )

This depends of course on what information you've got.
Edit: made a correction to the formula

[delton]

The formula W = ( \mu mg \cdot \vec{F}_\textrm{normal} ) ( AB + BC + CD + DA )does indeed work. But remember that work equals
force times displacement, not distance. The displacement (net distance traveled) in this case is zero, so the work is zero. If you want to show this, make CD and DA negative distances.

[gschjetne]

OK, I see. Thanks for clearing that up.

[Doc Al]

Hey, couldn't you just use this formula:

W = ( \mu mg \cdot \vec{F}_\textrm{normal} ) ( AB + BC + CD + DA )

This depends of course on what information you've got.
Edit: made a correction to the formula
What's this formula supposed to be? What do you mean by \mu mg \cdot \vec{F}_\textrm{normal} ?