# I I'm confused about net work

1. Nov 10, 2016

### RanaiD

I'm struggling with the concept of net work. If I had a box and was pushing it up an incline would the net work be all the forces parallel to displacement in x direction * displacement in the x direction added up, plus the displacement in the y direction * force gravity parallel to displacement in y direction. Or would I separate the net work based on whether displacement is in y and x direction.
In other words is net work the total work in both directions of displacement on an incline added up OR is the net work calculated separately for x direction and y direction displacement.

I ask this because I always thought that net work on an incline would be the work done in vertical direction displacement on incline as well as work done in horizontal direction, but when I solve problems that look for change in kinetic energy, change in k = net work and the only work included is in the direction of displacement.
I tried to reconcile this using 1/2mv1^2+mgh1=1/2mv2^2+mgh2
But I got:
1/2mv^2=mgh
Which is different from change in k=net work
1/2mv^2=Work in x direction

So I guess this more three questions than one.

Thank you for any help!

2. Nov 10, 2016

### Simon Bridge

Work is always done by something on something.

Net work done on something would usually be the net force on it multiplied by the distance it moves.

So net work for pushing an object up an incline would be the net force acting up the hill multiplied by the distance moved along the slope.
This would also be the change in total energy ... so the work done on the object would be: $W = mg(y_2-y_1) + \frac{1}{2}m(v^2-u^2)$ in the absence of friction.
With friction there is an extra term for whoever/whatever is doing the work against the friction. This will change the internal energy of the object so you have to decide if this should be included in the work done on the object. That will depend on the context.
That help?

3. Nov 10, 2016

### RanaiD

So does that mean you only look at the work of the forces acting parallel to the x displacement of the box moving up the incline?
I understand why you would add the work done by gravity along the horizontal displacement, but why wouldn't you also add the vertical displacement done by gravity, mgcos(theta)*y displacement

Because when I calculate the net work on a problem with a box being pushed up an incline I always want to do:
Wgx + Wpush+Wfr+Wgy= net work
Where wgx is the work done by gravity to push the box horizontally along the incline
And wgy is the work done by gravity on box in its vertical displacement up the incline.

Thank you so much again for your help!

4. Nov 10, 2016

### Simon Bridge

Force F and displacement s are vectors ... by definition, the work is $W=\vec F \cdot \vec s$ ... that is,it is the dot product of the net force with the displacement.

I did. But I think I didn't make myself clear - in my example formula using work = change in energy, I had implicitly defined the +y direction as "upwards", so $\vec F_{grav}= -mg\hat\jmath$. When you consider that the displacement is $\vec s = \Delta x \hat\imath + \Delta y\hat\jmath$ you can see how the dot product comes out.
The trick is to work out where the energy comes from and where it goes so you can make sense of the signs.

I think you need to take care to define what you mean by "net work" and what is doing it on what.
This can change depending on the context.

$W=\vec F_{net}\cdot\vec s$ would be the work done by the net force ... the work done on the box by the net force would be the negative of this.

For collisions - the "net work" is the average force during the impact times the distance moved during the impact.

Some other discussions:
http://physics.stackexchange.com/questions/189190/is-net-work-and-total-work-same

This one may be closer to your confusion:
http://physics.stackexchange.com/qu...en-we-lift-it-and-put-it-on-the-table-is-zero