Is the work caused by the force of friction negative?

[Jamesgortrig]

Homework Statement

A student pushes a 0.65kg box Ali g a desk. When he stops pushing the book, it moves 85cm before stoping (slowing down in this period). Coefficient of friction between book and Table is 0.27.Calculate the work done on the book by the friction. Should it be positive or negative?

Homework Equations

Doesn't matter. Just need theory explanation.

The Attempt at a Solution

I'm struggling on the last part. I calculated work to be 1.46J, but I don't know if it's positive or negative. I say it's. Negative, because the force of friction is opposite to the displacement, but the answer says positive?

 

[Jamesgortrig]

Sorry. It should say he pushes a book ALONG a desk.

 

[Simon Bridge]

I usually work this sort of thing out by considering what is doing the work on what.
Work done on the body is the change in that bodies energy ... which is negative if the bodies energy decreases.

Think gravity... when the body is lifted, there is negative work done by the gravitational field and positive work done by the lifting force.

That way you are thinking in phyiscal terms rather than just mathematical definitions.

BTW: By definition, whatever applies the force does work according to: $$\mathrm{d}W = \vec F \cdot \mathrm{d}\vec s$$

 

[atomicpedals]

To compliment the first reply - keep in mind the following: If the force has a component in the same direction as the displacement of the object, the force is doing positive work. If the force has a component in the direction opposite to the displacement, the force does negative work.

 

[PeroK]

I'm struggling on the last part. I calculated work to be 1.46J, but I don't know if it's positive or negative. I say it's. Negative, because the force of friction is opposite to the displacement, but the answer says positive?

I'd agree with you. The work done by friction is negative in this case, because a) it is in the opposite direction from the displacement and b) it reduces the KE of the book.

 

[Jamesgortrig]

My net energy statements are...
ΣE2 = ΣE1 - w

In this case, calculating a positive answer badges server right? But I would still consider it to be negative?

 

[jbriggs444]

My net energy statements are...
ΣE2 = ΣE1 - w

The sum of all energies afterward equals the sum of all energies before minus work done in between?
Work done by external entities on the system?
Or work done by the system on external entities?

 

[Jamesgortrig]

The sum of all energies afterward equals the sum of all energies before minus work done in between?
Work done by external entities on the system?
Or work done by the system on external entities?

Work done on the book by the table (resulting in the friction). So technically external because it's a normal external interaction with the desk

 

[jbriggs444]

Work done on the book by the table (resulting in the friction). So technically external because it's a normal external interaction with the desk

Why would you subtract the energy that is added to the system?

 

[Jamesgortrig]

Why would you subtract the energy that is added to the system?

Well, in the first instant of time when the book is being pushed, there's more energy than the second instant. I'm subtracting because friction is in opposite direction of displacement. With the force pushing the book gone, ∑E2 should have the work subtracted from it, as the friction from the table is working against the book.

 

[jbriggs444]

Well, in the first instant of time when the book is being pushed, there's more energy than the second instant.

Stop right there.

If you are talking about the book being pushed then the external work being done is from the push. Friction is not yet being considered. So do not consider it.

The work that is being done by the push should be added to the initial energy of the book to get the final energy of the book.

Now we can proceed to consider friction after the push has ended. The work done by friction on the book is still added to the kinetic energy of the book. If that work is positive then adding it to the initial kinetic energy results in an increase. If that work is negative then adding it to the initial kinetic energy results in a decrease.

In no case do you subtract the work done on a system from the energy already in the system. That's a sign error.

 

[Jamesgortrig]

Stop right there.

If you are talking about the book being pushed then the external work being done is from the push. Friction is not yet being considered. So do not consider it.

The work that is being done by the push should be added to the initial energy of the book to get the final energy of the book.

Now we can proceed to consider friction after the push has ended. The work done by friction on the book is still added to the kinetic energy of the book. If that work is positive then adding it to the initial kinetic energy results in an increase. If that work is negative then adding it to the initial kinetic energy results in a decrease.

In no case do you subtract the work done on a system from the energy already in the system. That's a sign error.

So under no circumstances I subtract work in an energy statement? It's always added?

 

[jbriggs444]

So under no circumstances I subtract work in an energy statement? It's always added?

You could subtract the work done on a system from the final energy of that system to get the initial energy of that system.

You could subtract the work done by a system from the initial energy of that system to get the final energy of that system -- if you had an assurance that there were no dissipative losses. [But with sliding friction there will be dissipative losses, so that's a non-starter in this case]

 

[vela]

So under no circumstances I subtract work in an energy statement? It's always added?

If you're using the conventional definition of work, you should always have $E_\text{f}-E_\text{i}=W$. What you don't want to do is say that because friction is going to slow the object down that $E_\text{f}-E_\text{i}=-W$. In the case of kinetic friction, the work is negative ($W<0$), so putting in the minus sign into the equation by hand is unnecessary.