- #1
RanaiD
- 3
- 0
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!
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!