Why is total work 0 on U ramp?

Leypoldon

Leypoldon

Open question

2022-08-19

Suppose there is a U shaped frictionless skateboard ramp. If I put on far top left of the ramp and then release it until it reaches top left then from Δ K = W t o t = 0 because initial and final velocity is 0. I am quite confused that total work is zero. When I think in terms of W t o t = a b F n e t d r I think the total work should be non zero because it seems that the force is always inclined with the dr direction. Why does it seems to be contradicting each other?​

Answer & Explanation

Alexia Mata

Alexia Mata

Beginner2022-08-20Added 15 answers

The force of gravity always points downward. Along the path y = x 2 , for instance, F d l is positive when the ball is on its way down and negative on its way back up. This cancels and the overall work done on the object is 0. To see this, let
F = α y ^
In Cartesian coordinates
d l = d x x ^ + d y y ^
The dot product is
F d l = α d y
Along our path (P)
d y = 2 x d x
So that, finally
P F d l = 2 α 1 1 x d x = 0
Of course, there's another force keeping the object moving along our path, but the work done by it is 0. This can be taken as a given, since the ramp is frictionless, or it can be worked out a bit more rigorously, similarly to the above.

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