Why is there potential Energy due to lifting an object or moving when it's attached to a spring while the net work is zero?

Consider that you lift an object a height h vertically, if the body happens to move with uniform velocity this means that net force acting on it is zero.

The same is true for a body attached to a spring, I apply a force in opposite direction to that of the string (They vary with each other) so they cancel out and hence he body moves uniformly.

So in the above two examples, the way I understand it, the net work is zero and the work is change in Energy, so if there's no change in Energy how come there's Potential Energy?

Thank you!

PS: I've seen this question being asked before, but unfortunately I wasn't answered, so if you may please make the answers as simple as possible, and from other prespectives.

• What force makes the body move uniformly? Take that into account and you'll have your "missing" energy. – ACuriousMind Dec 28 '15 at 12:28
• If it moves uniformly then there are no forces acting on it. – Raafat A. Ali Dec 28 '15 at 12:48
• It sounds like you're trading gravitational potential energy of the object with elastic potential energy of the spring, for a zero net potential energy change. – James Dec 28 '15 at 13:31
• If it moves uniformly against a gravitation field (as it has to if it is gaining potential energy) there has to be a force acting on it that exactly counters the gravitational force. – ACuriousMind Dec 28 '15 at 16:05

How do you lift a body up from the ground?

The body remains at equilibrium under the acting of opposite forces on it: the $\uparrow$ normal force from the ground & the $\downarrow$ gravitational force . In order to move it up, you have to give some infinitesimal $\uparrow$ force greater than gravitational force which would accelerate the body upwards.

Let the force be $\mathbf F_\text{ext}.$ Therefore the equation of motion provides:

$$\mathbf F_\text{ext}- m\mathbf g = m\;\delta \mathbf a\implies \mathbf F_\text{ext}= m\mathbf g + m\; \delta \mathbf a \;.$$

Let the body gets displaced above by $\bf h\;.$ Therefore the work done by $\mathbf F_\text{ext}$ is given by \begin{align}W_{\mathbf F_\text{ext}}&= \mathbf F_\text{ext}\cdot \mathbf h\\ &= m\mathbf {g}\;\cdot \mathbf h + m\; \delta \mathbf a \;\cdot\mathbf h \;.\end{align}

This work can be assessed as $$\overbrace{m\mathbf {g}\; \cdot\mathbf h}^{\text{work done against gravity}} + \overbrace{m\; \delta \mathbf a \;\cdot\mathbf h}^{\text{amount of kinetic energy increased}}\;.$$

The former work is what you call potential energy- the work done against the gravitational force. Hadn't the force worked against the gravitational force, then the body would have been constantly decelerated by the gravitational force turning all the kinetic energy into potential energy of the system.

At the present context, the later work is zero; so the body is moving at constant velocity.

But how?

There must be a force $\mathbf F_\text{ext}$ to counteract the gravitational force; otherwise the body couldn't move at constant velocity as its kinetic energy would have been converted to potential energy of the system. To counteract the work done on the body by the gravitational force, same amount of work is done by the external force on the body so as to retain its constant velocity; that work turns up as gravitational potential energy.

So, although the net work $W_{\mathbf{g}} + W_{\mathbf F_\text{ext}}= -m\mathbf g\;\cdot \mathbf h+m\mathbf g\;\cdot \mathbf h= 0 \;,$ in order for the body to move at constant velocity, the external force must have to do the same amount of work as done by the gravitational force- this is the potential energy of the system.

You're examining the net work done on, say, a block attached to the end of the spring. Indeed, if you consider the met force on the block while it travels at constant velocity, you calculate that no work has been done on the block.

Considering the entire system though, your hand pushing the block applies a force over a distance and the spring also applies a force over that same distance, opposite to the force from your hand. We can actually ignore the block here, since it is inert, and say that your hand does work on the spring and vice versa.

Take two cases, one where the spring is anchored and one where it is not. In the unanchored case, a certain amount of force accelerates the spring to a certain velocity and it floats away through space. If the spring is anchored, the same work does not accelerate the spring: it displaces one end, but the final velocity is still zero. Your hand imparts the same energy to the spring in both cases, but in one case the result is kinetic energy of the spring, and in the other case the result is potential energy stored in the spring structure. If you place the block in front of the compressed anchored spring and move your hand away, the block will be fired away with the same momentum as the spring would have gained if you'd pushed it without anchoring it first.

So yes, in your original example the net work on the block is zero, but the total energy of the system is not zero, and the total work is also not zero, although it is conserved.

net work is zero and the work is change in Energy

actually change in the kinetic energy is equal to the work done. You when you move an object vertically without any acceleration then there no change in the kinetic energy hence from the work-energy theorem the work done is zero.

You understand the Potential Energy in this way: if an object is having potential energy then it has some "potential", some energy, that will make it to do work if other forces are removed like in the case of object being pushed vertically upward. Suppose you pulled that mass upto some height and then hold it, you would be applying a force which is equal to mass' weight but opposite in direction, however if you droped that object, you would have stop applying the force which was holding that mass up, then it would fall down, i.e. it will do work as it falls, but why it does that work? to explain this the concept of potential energy was introduce to explain shuch type of phenomenon.