Why Does Gravity Do Equal Work on All Inclines?

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Bashyboy
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Hi,

Currently I am reading about work, and the author of an article I am reading states that "You can
push an object up an incline and the amount of work done by gravity is the same for all angles
of the incline." Why is this so?
 
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Because gravitational work depends only on the difference in altitude between start and finish, not the route taken.

An intuitive way to look at it is that it's hard to push a weight up a steep slope, but it's not far to go. It's easy to push a weight up a gentle slope, but you will be going a lot further. The change in "how hard it is" and in" how far it is" cancel out as you change the angle of the slope.
 
What's the direction of the gravitational force? Only the component of the displacement parallel to gravity will count towards that work.
 
in an ideal situation with no friction in a gravitational setting, moving an object from side to side no work is needed ie the object's potential energy doesn't change it doesn't slide back to the original spot. However moving it vertically does require work and when you let go, the object moves back to its original position. So for an incline it doesn't matter what angle it is, what counts is how much the object moves vertically as to how much work is required and how much potential energy it gains.
 
Bashyboy said:
Hi,

Currently I am reading about work, and the author of an article I am reading states that "You can
push an object up an incline and the amount of work done by gravity is the same for all angles
of the incline." Why is this so?

One of the consequences of the formal definition of work that bothered me a lot when I first saw it (a LONG time ago) was that if you pick up a heavy rock, carry it up a hilll and then carry it back down the hill and put it back where you found it, the total work you have done is ZERO. The math may say that, but your back will tell you another story. :smile: