# Why is total work 0 on U ramp?

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 $$\Delta K= W_{tot}=0$$ because initial and final velocity is 0. I am quite confused that total work is zero. When I think in terms of $$W_{tot}=\int_a^b \overrightarrow{F}_{net}\cdot \overrightarrow{dr}$$ 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? ​

• "If I put on far top left of the ramp and then release it until it reaches top left" Don't you mean until it reaches the top right? Mar 5, 2022 at 18:40

The force of gravity always points downward. Along the path $$y=x^2$$ on $$-1, for instance, $$\mathbf{F} \cdot d\mathbf{l}$$ is positive when the ball is on its way down and negative on its way back up. This cancels and the overall work done on the object is 0. To see this, let $$\mathbf{F}=-\alpha\mathbf{\hat{y}}$$

In Cartesian coordinates $$d\mathbf{l}=dx\mathbf{\hat x} + dy\mathbf{\hat y}$$

The dot product is $$\mathbf{F}\cdot d\mathbf{l}=-\alpha dy$$

Along our path (P) $$dy=2xdx$$

So that, finally $$\int_{P}\mathbf{F}\cdot d\mathbf{l}=-2\alpha\int_{-1}^{1}xdx=0$$

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.

Neglecting all forms of friction, there are only two forces acting on the object that are responsible for doing work: Gravity and the reaction force of the ramp.

The reaction force of the ramp is responsible for horizontal displacement of the object. But since the ramp is vertically symmetrical, the horizontal displacements due to the ramp forces are equal and opposite and thus cancel out for a net work of zero by the ramp.

Gravity is a conservative force so that its work depends only on the vertical displacement of the object. When the object is released on the left side gravity does positive work, since the direction of its force is the same as the displacement of the object, resulting in $$\Delta KE=+\frac{1}{2}mv^2$$ at the bottom of the ramp. For conservation of energy, the gain in KE equals the loss of GPE, or $$\Delta GPE=-mgh$$.

As the object moves up the right side of the gravity now does negative work since its force is in the opposite direction of the displacement of the object. When something does negative work on an object it takes energy away from the object. In this case gravity takes the KE away from the object and stores it as GPE. From the bottom of the ramp to the top right side when the object once again comes to rest, $$\Delta KE=-\frac{1}{2}mv_{2}$$ and $$\Delta GPE=+mgh$$,

Per the work energy theorem, the net work done on an object from the upper left side to upper right side of the ramp equals its change in kinetic energy. In this case the positive work done by gravity on the left side of the ramp equals the negative work done by gravity on the right side of the ramp, for a net work of zero and $$\Delta KE=+\frac{1}{2}mv^{2}-\frac{1}{2}mv^2=0$$, per the work energy theorem.

Hope this helps.