Is Net Work Zero If Force is Applied in Opposite Directions?

[17_]

This may seem like a silly question, but I just want to make sure my thinking is correct.

If you push a box, let's say 5m with 50N of force then back to where it started with 100N of force, is the net work done zero? I mean sure the work you do going back is more than the work you put into get it there, but the total displacement in the end is zero.
Thanks

 

[Bystander]

Nope. Work doesn't care which direction you're going. It's a scalar.

 

[17_]

But isn't the total displacement zero? So therefore work would be zero?

 

[Bystander]

Scalar. You can push it around in a circle and return to the same place. You have still done work. Push it down the hill against friction on a ramp with gravity helping, and back up the hill against friction and against gravity, and you've done work both directions. Neither process suddenly puts fuel back into the bulldozer's tank when you return to the original position.

 

[jbriggs444]

But isn't the total displacement zero? So therefore work would be zero?

As long as the force is constant and always in the same direction then work can be easily computed as force times total displacement in the direction parallel to the force. In that case, a zero total displacement would mean zero total work.

If you push a box somewhere and then push it back, the force is not always in the same direction. You need to compute the work done pushing the box out and add it to the work done pushing the box back separately and add those results together to determine the total work done.

More generally, work is the "path integral" of instantaneous force times incremental distance.

 

[jtbell]

Mentor's note: Abdulrehman's homework question and its responses have been moved to a new thread here:

https://www.physicsforums.com/threads/quick-question-about-work.784561/