Work & Energy Transfer: Comparing Configurations

[Perrin]

Hello, I've been going over work and energy transfer for some upcoming tests, and I got the following question:

http://www.dotcore.co.il/conf1.gif

If I look at that configuration, the ball is moving down the inclined path in the same direction as the x axis, and there's no movement on the y axis, so the only force doing work is the gravitational force. The work being done I calculated as:
$$W=mg\cos{\Theta}\frac{h}{cos{\Theta}}$$
Which I simplified as:
$$W=mgh$$

Now, that tells me that the gravitational force is doing work on the ball equal to mgh.
If I setup a different configuration though:

http://www.dotcore.co.il/conf2.gif

Now, the gravitational force is doing work only in the y axis, equal to the work it did in the last setup, but the normal force is doing work on the x axis, equal to:
$$W=N\cos{\Theta}h\tan{\Theta}$$
$$W=Nh\sin{\Theta}$$

Now, does this mean that there is more work done in one setup than in the other? Did I make a mistake?
Thanks for the help!

 

[Doc Al]

Now, the gravitational force is doing work only in the y axis, equal to the work it did in the last setup, but the normal force is doing work on the x axis, equal to:
$$W=N\cos{\Theta}h\tan{\Theta}$$
$$W=Nh\sin{\Theta}$$

Now, does this mean that there is more work done in one setup than in the other? Did I make a mistake?

Don't forget the work done by the vertical component of normal force.

(Since the displacement is perpendicular to the normal force, you know that the net work done by the normal force must be zero, no matter how you define your coordinate system. )

 

[Perrin]

Ohhh, so that's what I forgot! It's doing negative work equal to Nhsin(theta).

Thanks for clarifying that! I think I got a better grip on this now