A student lifts a 20.0 kg box 1.5 m straight up at a constant speed. Find the work done by each force that acts on the box.

Question

anonymous · Answer

I tried it and got (20)(9.8)(1.5) for Applied Force- would it just be negative for Fg?

roadjester · Answer

$\large W=F\Delta x cos\theta$

roadjester · Answer

I'm actually confusing myself...
The applied force is upward due to lifting upward but gravitational acceleration is downward...
@theEric

roadjester · Answer

WAIT, acceleration is 0! there is no work upward since it's constant velocity... I think...

theeric · Answer

There is no work done on the box, but that's because the work done by the lifter and the work done by gravity are equal in magnitude and opposite direction. So they cancel!

$W=Fd\cos\theta$ where $\theta$ is the angle between the force and displacement.

anonymous · Answer

So I was right then? The work done by Fg is just negative work done by Fa?

theeric · Answer

@greenglasses You are right!

theeric · Answer

Yep!

roadjester · Answer

Thanks @theEric I was confusing the heck out of myself

anonymous · Answer

Thanks, my teacher didn't leave us any answers so I had no way to check.

theeric · Answer

Haha, no problem. I dearly hope I'm not confused! :)

theeric · Answer

Cool! I'm glad OpenStudy is here.