Work Done going up an incline:

[Peter G.]

I read in my book that when we are calculating the amount of work done by a man to go up a flight of stairs we use as force his weight and the distance, the vertical displacement.

Then, looking at this website http://www.physicsclassroom.com/mmedia/energy/au.cfm they illustrate a car going uphill through three different paths, each differing in steepness. The steepest one requires the most force but covers a shorter difference. The less steep one requires less force but covers a greater difference. I understand that they all end up doing the same work because they all end up at the same height. But, in this example, unlike in the staircase, they don't use the vertical displacement, instead they use the displacement through the path, why?

Thanks
Peter G.

 

[SammyS]

Do you know trig?

Your book takes the component of displacement parallel to the applied force (Gravity is down, so the applied force is up.) times the applied force. This is the same as taking (magnitude of the displacement) × (cosine of angle between displacement & applied force) × (magnitude of the applied force).

The website takes the component of the force needed to overcome gravity that's parallel to the displacement times the displacement. The website only looks at the component of the force parallel to the road. That is (magnitude of the weight) × (cosine angle between the displacement and the up direction - which is opposite gravity). Then they take this times the magnitude of the displacement.

So, they're the same.

 

[Peter G.]

I drew, did the maths and got it! Thanks a lot