Question

Two closed systems.
System x:
F = 450N
m = 25g


System y:
F = 225
Mass:6g


Both systems are in motion
What would have more energy?
System x/y?

Answer 1

@experimentX @Jemurray3 @TuringTest @Mashy

Answer 2

More *Kinetic Energy*

Answer 3

Both are in motion.

Answer 4

That would clearly depend on how fast they were moving.

Answer 5

I'd say that x would have low acceleration compared to y because of the mass.

Answer 6

Right, but energy doesn't care about acceleration, it cares about velocity, so that's irrelevant.

Answer 7

I guess... V in x would be greater due to the force?
I mean since y has less F...

Answer 8

If V is not given we can't figure it out from those variables alone?

Answer 9

or at least guess.

Answer 10

No, you can't. You're still struggling with the idea of force, I think.

Answer 11

Kinda yea.
But that simply means I'd have to measure the velocity.
Then figure the out the answer.

Answer 12

However, the reason I said that system X has more energy.
is say the distance between them are equal. Lets say distance is = to 1M
W(x) = 450x1
W(y) = 225 x 1
Since both systems share the same area of space the force is going to be applied over... I can figure the work done, and compare their energy.

Answer 13

The reason for this question.
Is my confusion of mass.
I assume having more mass in a system = more energy
Less mass = less energy.

Bu that assumption is flawed without the proper calculations.

Answer 14

By the distance between them being equal, I assume you mean that the forces applied are equal. In that case, you can find the energies of the systems, and it would turn out that system x would have more energy.

The point I have been trying to get across is that a force is just an influence exerted on something. To find out how much energy it imparts to the system, you need to know the distance over which it is applied.

Answer 15

A much simpler calculation is that the net work done on a system is equal to its change in energy.

Answer 16

kinetic energy*

Answer 17

E = W = Fd = KE = 1/2mv^2
I assume that is what you ment?

Answer 18

Forgot delta... But hey!
I hope you got the picture.

Answer 19

Assuming that you're not worrying about potential energy, the force is parallel to the displacement, and it wasn't moving beforehand, all those equal signs are fine...

Answer 20

It's better to just say
\[ W = \Delta (KE) \]

Answer 21

loool yea.

Answer 22

Or, being more precise,
\[W_{net} = \Delta(KE) \]

Answer 23

"By the distance between them being equal, I assume you mean that the forces applied are equal. "
I ment that, both systems applied their "different" forces in the same distance.

Answer 24

Yeah, that's what I meant to say, sorry.

Answer 25

+ I would usually use this:
\[W _{net}^{} = \Delta(KE)\]
When I have all the variables. But in some cases like this...
When I can't, and I need a quick guess and only have the distance and force
I'll use the old classic:
\[W _{net}^{} = Fd\]

Answer 26

You are misunderstanding me. The net work is equal to the change in kinetic energy, always, no matter what. If you can calculate the work, then that's the change in kinetic energy.

Answer 27

If its rotational energy that is required... I use this formula:
\[W _{net}^{} = \tau \Theta \]

Answer 28

Ow ok ok! Thanks.

Answer 29

lol, sometimes I forget the small details!

Answer 30

@Jemurray3 Thanks again buddy I can always count on ya!

Answer 31

No problem.

Answer 32

Take care!