Relation between sum of the forces and energy

[Karagoz]

In the picture above, there are three balls in separated small elevators. Elevator A lifts the ball upwards, elevator B stays still, elevator C moves the ball downwards, all in constant speed. (And this is a model, a simplification of the reality, we assume no other forces on the balls other than gravity force downwards and normal force upwards).

Since there's no acceleration, only constant speed or staying still, sum of the forces on the ball is 0 in both A, B and C.

But lifting the ball upwards requires more energy than keeping the ball still. But how is it that sum of the forces on the ball is equal (and zero) both in A, B and C?

If you have a force meter, and hang a 1KG object on it. Won't the force meter show higher value when you move the force meter (with the 1KG object hang on it) upwards in constant speed, than when you keep it still?

 

[haruspex]

lifting the ball upwards requires more energy than keeping the ball still. But how is it that sum of the forces on the ball is equal (and zero) both in A, B and C?

Yes, work done by a force is the force multiplied by the distance the force advances. In A, the normal force does work on the ball, while gravity does equally negative work. Result, no speed change.

Won't the force meter show higher value when you move the force meter (with the 1KG object hang on it) upwards in constant speed, than when you keep it still?

No, for the reasons you gave.

 

[Karagoz]

Yes, work done by a force is the force multiplied by the distance the force advances. In A, the normal force does work on the ball, while gravity does equally negative work. Result, no speed change.

If we assume the ball is 1KGs, then both in A, B and C the normal force to the ball is ca 9.8N?

When the forces are the same, then how is it that the ball in A moves upward (and does more work), but the ball in B is staying still?

 

[Dale]

But lifting the ball upwards requires more energy than keeping the ball still.

Be careful here. Think about what exactly you mean by "requires more energy"

 

[russ_watters]

When the forces are the same, then how is it that the ball in A moves upward (and does more work), but the ball in B is staying still?

You tell us: it's your scenario. What element of the scenario do you have in mind that you didn't tell us about? I'm guessing you described step 3 but didn't tell us what you want for steps 1 and 2...

 

[Karagoz]

Be careful here. Think about what exactly you mean by "requires more energy"

I mean that it consumes more energy to move something upward.

Let's assume the elevators in the picture are electrical.

If the elevator is at the bottom and staying still there (like in B), then we don't need any electricity to keep it there.

But if the elevator is moving upward (like in A), and to keep it moving upward with constant speed, we need to consume electrical energy.

In A we need to consume electrical energy, but not in B.

But both in A and B, the forces on the ball and sum of these forces are the same, all zero.
Why is it that way?

Or see this short experiment with weight scale on elevator.


It shows 122. Then it increases to 130 when accelerating. But then goes back to 122 when the elevator escalates upwards.

Why it's not in 130 all the way when the elevator is escalating upwards?

 

[jbriggs444]

Why it's not in 130 all the way when the elevator is escalating upwards?

Because, after the initial upward jerk, there is no acceleration and, accordingly, no net force. The elevator must be providing a force exactly equal to that of gravity.

The concept of "work" which is computed as force times [parallel] distance moved is useful here -- it determines the energy supplied by that force.

 

[haruspex]

When the forces are the same, then how is it that the ball in A moves upward (and does more work), but the ball in B is staying still

There is no contradiction. It defies intuition because you are not accustomed to forces remaining constant when being allowed to perform work. Mostly, when you take off a brake that had been preventing a force from doing work the magnitude of the force declines somewhat.
Similarly, your experience is that moving a mass has to overcome friction, so it takes a greater force to sustain motion than merely to hold things in place.
But in principle, two constant equal and opposite forces on a mass will produce no acceleration, so the mass will move at constant velocity. If the motion has a component parallel to the forces then one force is doing work against the other.
Consider e.g. two equal masses connected by a string over a frictionless beam. They may hang in equilibrium, but if you provide an initial uoward push on one then they would keep moving at that speed, the descending mass losing PE and the other gaining it.

 

[russ_watters]

It shows 122. Then it increases to 130 when accelerating. But then goes back to 122 when the elevator escalates upwards.

Why it's not in 130 all the way when the elevator is escalating upwards?

That's Step 2, that you omitted from your scenario. You gave us Step 3, which is moving upwards at constant speed. You didn't describe but are obviously concerned about Step 2. Step 1, evidently, is sitting stationary. So how does an object go from stationary to moving at constant speed? That's what Step 2 is.

 

[Dale]

I mean that it consumes more energy to move something upward.

Let's assume the elevators in the picture are electrical.

Note that "it" in your first sentence refers to "the elevators" in your second sentence. So the only energy that you are interested in is the energy required by the elevator mechanism. So the difference between the scenarios is not the magnitude of the forces, but the direction of the displacement. In one case the displacement is in the same direction as the force from the elevator mechanism. In another case it is in the opposite direction, and in the other case there is no displacement. The net force is not relevant to the work done by the elevator mechanism. Only the elevator mechanism's force and the displacement

 

[Karagoz]

There is no contradiction. It defies intuition because you are not accustomed to forces remaining constant when being allowed to perform work. Mostly, when you take off a brake that had been preventing a force from doing work the magnitude of the force declines somewhat.
Similarly, your experience is that moving a mass has to overcome friction, so it takes a greater force to sustain motion than merely to hold things in place.
But in principle, two constant equal and opposite forces on a mass will produce no acceleration, so the mass will move at constant velocity. If the motion has a component parallel to the forces then one force is doing work against the other.
Consider e.g. two equal masses connected by a string over a frictionless beam. They may hang in equilibrium, but if you provide an initial uoward push on one then they would keep moving at that speed, the descending mass losing PE and the other gaining it.

Is this what you mean?

Note that "it" in your first sentence refers to "the elevators" in your second sentence. So the only energy that you are interested in is the energy required by the elevator mechanism. So the difference between the scenarios is not the magnitude of the forces, but the direction of the displacement. In one case the displacement is in the same direction as the force from the elevator mechanism. In another case it is in the opposite direction, and in the other case there is no displacement. The net force is not relevant to the work done by the elevator mechanism. Only the elevator mechanism's force and the displacement

So it's the elevator mechanism that uses energy to lift the ball against the gravity in constant speed.

But when the elevator is still at the ground, the elevator mechanism doesn't need to use energy to keep it there. Still there are some "normal force" on the elevator so the elevator is not pulled down by the gravity. But what's the source of the energy of that "normal force"?

 

[jbriggs444]

But what's the source of the energy of that "normal force"?

The point of action of that normal force does not move. So it does no work and requires no energy.

 

[Dale]

But when the elevator is still at the ground ... what's the source of the energy of that "normal force"?

The displacement is 0 so no work is done and no energy is required.

 

[haruspex]

Is this what you mean?

Yes.