# Power and Change of Kinetic Energy in Variable Mass System

1. Nov 23, 2015

### PeteSeeger

I am thinking about something that is getting me quite confused. To illustrate what is confusing me let's have a cart moving in the horizontal direction on a track without friction, and a motor which adjusts to keep the kart at a constant speed V in the positive direction. Now, it's snowing so the mass of the cart is increasing at a rate dm/dt. The power the motor must do to keep the cart going at a constant speed is P=(dm/dt)V2. But, the change in kinetic energy is dT/dt=d((1/2)mV2)/dt=(1/2)(dm/dt)V2. So, if we integrate the power and change of kinetic energy over some interval of time, we find that the interval of work by the motor is twice the change in kinetic energy. Where is the other half of the work going? There is no friction in the problem. I understand why mathematically it doesn't work out as usual where these things are equal (because W≠ΔT in this problem because the integral of force doesn't work out that way because the change in mass), but I am struggling to conceptually understand this scenario.
(I should emphasize the snow is falling with no horizontal velocity relative to the ground, and the motor is keeping the cart at a constant velocity since it is inclined to slow because of conservation of momentum).

Last edited: Nov 23, 2015
2. Nov 23, 2015

### PeteSeeger

I have thought about this more and I will give a theory of mine. When the snow lands in the cart, the second the cart contacts the snow it exerts a force to speed it up. At the same time, the snow exerts a force back trying to slow down the cart. So, the cart must overcome the equal and opposite force to stay at a constant velocity, as well as accelerate the snow. Does this correctly explain why the change in kinetic energy is one half the work?

3. Nov 23, 2015

### UncertaintyAjay

No. I dont think that's it.

Work Done is still equal to change in kinetic energy. What would that work be used for if not to increase kinetic energy? As you said work isn't lost at heat.
About the discrepancy, this is what I think:
The formula K.E= 1/2 mv2 is derived for the case where mass is constant . I don't think it applies in this scenario. What does apply in this case , and in most others, is W=K.E. I can derive the K.E = 1/2 mv2 for you, but maybe you'd like to try it yourself and see how that assumes mass to be constant. A formula derived under certain conditions or assumptions is only valid if those assumptions are still true. They aren't in this case.
Anyway, that's my way of looking at it. It makes sense to me, but I'd like it if someone else could weigh in, see what they think of this explanation and the problem( good question). I'm probably about the same age as you,so I'm not the highest authority on this.

4. Nov 23, 2015

### PeteSeeger

Thanks for the reply. I thought kinetic energy was a defined to by (1/2)mv2, so that in this case work wouldn't be simply the change in kinetic energy, but is it instead defined to be the integral of force (so that work will always be the difference of kinetic energy)? That would resolve things.

5. Nov 23, 2015

### UncertaintyAjay

Yeah, it isnt defined as 1/2mv2

6. Nov 23, 2015

### Staff: Mentor

Where did that expression come from?

7. Nov 23, 2015

### PeteSeeger

P=Fv, and F=(dm/dt)V, so P=(dm/dt)V2

8. Nov 24, 2015

### Staff: Mentor

Some of the responses above are incorrect: the kinetic energy of a moving system most certainly is $mv^2/2$; the non-constant mass doesn't change this. And as for where the other half of the work (which you have calculated correctly) is going....

Consider the system as if you were moving with the cart. Now you are at rest while the snow is moving towards you with speed $v$. The kinetic energy of the cart is zero and it stays that way; no work is being done on the cart. But what happens to the kinetic energy of the moving snow when it strikes the cart and sticks to it, so now is also at rest? That's easy - the snow-cart collision is inelastic, so the kinetic energy is dissipated as heat during the collision.

Now when we shift back to thinking of the snow and the ground as being at rest and the cart moving, we see where the work is going. Half of it is spent accelerating the snow from zero to the speed of the cart, and the other half goes into waste heat from the inelastic cart-snow collision.

(Note that if instead of sticky snow we had rubber balls that rebounded elastically off the cart, they'd end up moving even faster than the cart and the "missing" work would be in their kinetic energy).

9. Nov 24, 2015

### UncertaintyAjay

Ahhhh. Thanks. I was just clutching at straws for an explanation. That question perplexed me too. Thanks for answering .

10. Nov 24, 2015

### PeteSeeger

Ah, ok, now I understand. Thanks everyone.

11. Feb 4, 2016

### Merlin the magic pig

This explanation works fine for a qualitative answer but why does it come out to be exactly half? Can this be shown algerbraically?