# Homework Help: Work and Energy

1. Jan 19, 2013

### quicksilver123

I was watching a video at Khan Academy on Work & Energy (link above).

At 2:50, he describes a situation with an elevator doing work against gravity.

My question is pretty simple:

If the acceleration (and hence, net force) is equal to zero (the upwards force cancelling out the downwards force due to gravity), how did the upwards force produce any movement at all?

The constant velocity means there's no acceleration, and that's fine.

But shouldn't there be no movement at all?

eg. [let down be negative]
If I hold my hand out and apply an upwards force of F=mg, wouldn't that merely counteract downwards acceleration due to gravity (F=-mg)? Wouldn't my hand merely stay still in the vertical plane?
Or, in a similar example more related to the one in the video, if my hand was held out and accelerating upwards due to a force, and that force changed to be equal and opposite to the force of gravity, would my hand continue to move upwards (with zero acceleration) at a constant velocity?

This is a conceptual problem that's been bugging me... even though the math supports what coursework teachers. I need help grasping this concept.

Last edited: Jan 19, 2013
2. Jan 19, 2013

### SammyS

Staff Emeritus
Once movement upwards is established, it takes a net force of zero to maintain the upward motion at a constant velocity. Newton's first law should tell you this. His second law also confirms this.

3. Jan 19, 2013

### quicksilver123

Ah. Simple enough - inertia.

It would take a downwards force greater than gravity to cause my hand to decelerate to a velocity of zero (at which point it would begin to fall).

Thanks. I know the laws work in math (and therefore in nature). I guess its just a matter of thinking of a variety examples until those laws are ingrained in my mind.

4. Jan 19, 2013

### quicksilver123

I know these are stupid questions - I'm just having trouble wrapping my head around some of these core concepts... and I need to master them before moving on.

A similar question - maybe someone can give me an example or something.

Work = displacement times force
w=df

(let the direction of motion be positive)

Let's say an object is sliding across my desk. How it started sliding doesn't matter.
It experiences negative acceleration due to kinetic friction.

Since f=ma
w=dma

Would negative work have been done, since acceleration is negative? Or are these absolute values?

Another one -
If I were to move an object (m=0.1kg) across my desk 0.5m with my hand at a constant speed (a=0), would zero work be done? After all, since a=0, w=0 as well... no?

5. Jan 19, 2013

### PhanthomJay

work, a scalar quantity, can be positive or negative, and is defined as the dot product of the force and displacement vectors, (f)(d)(costheta), where theta is the angle between the 2 vectors. In this case, the displacement is rightward and the force is leftward, so theta is 180 degrees, and hence, negative work is done by the friction force. Work done by a force does not necessarily require acceleration, however, as noted below in your next question.
The work done by the NET force is 0, Since the NET force is 0. So you can say that the total or net work done on the object by all forces acting on it is 0. But the pushing force of your hand on the object does positive work , and the force of the equal and opposite friction force acting on the object does negative work of the same magnitude, so together, the sum total of the work done on the object is 0.