Potential Energy and energy conservation

[kyin01]

Hi everyone, so I was going to start on my HW but there was some things confusing me when reading the book.
This is calculus based (feel free to use integrals) and we are on the concept of potential and total mechanical energy and such.1) This new chapter we are on introduces
Work_{grav}=mgh_{1} -mgh_{2} (1 being above 2) and

Work_{el}=.5kx_{1} ^{2} - .5kx_{2} ^{2}

However in the previous chapter we were told that W_{net}=\DeltaK_{energy} or W=F\bullet _{net} \vec{s}

So my question is, if the Work introduced now the same kind of work as we were told in the previous chapter?
I tried setting an quick example, let's assume we throw a ball of 10kg vertically upwards and it reaches a maximum height of 5m and than goes back down. Gravity is 10\frac{m}{s^{2}}. So let's pretend no air resistance and such and we are trying to find work.
If I use this equation W=F_{net}\bullet\vec{s}, I get work is = to 50J
but if I use the new equation Work_{grav}=mgh_{1} -mgh_{2}, work is = to 500

So what is it I am not understanding? Are they 2 different kind of work?2) Is internal energy (U_{int}) the same as work done by non conservative forces (W_{nc})?

edit: i have no idea why the subscripts appear as superscripts, I tried editing and fix but it still appears as superscripts. If it's not clear enough please let me know I'll make them into pictures in MSPaint and post or something

 

[Doc Al]

So my question is, if the Work introduced now the same kind of work as we were told in the previous chapter?

Sure.

I tried setting an quick example, let's assume we throw a ball of 10kg vertically upwards and it reaches a maximum height of 5m and than goes back down. Gravity is 10\frac{m}{s^{2}}. So let's pretend no air resistance and such and we are trying to find work.

If the ball goes up and down, the net work done by gravity is zero. (The work is negative on the way up, positive on the way down.)

Let's say you want the work done by gravity when a ball falls from a height of 5m to the ground, which is what I think you were calculating.

If I use this equation W=F_{net}\bullet\vec{s}, I get work is = to 50J

How did you get this? The force = mg = 100N. If we call the distance h=5m, the work done = mgh = (100)*5 = 500 J.

but if I use the new equation Work_{grav}=mgh_{1} -mgh_{2}, work is = to 500

This also gives you mgh = 500 J.

 

[kyin01]

Ahh, sorry I see it now. Thanks

So is Is internal energy the same as work done by non conservative forces?