Work-Energy Theorem question with not much detail to work with...

tbyrne

1. The problem statement, all variables and given/known data
Suppose a hockey puck of mass m is at rest on the ice. A surly Canadian hits the puck and sends it sailing across the ice at velocity v. According to the work-energy theorem, how much work did the player's stick do on the puck?

Doc Al

Well, what does the Work-Energy theorem state?

tbyrne

It states "the net work done by all the forces acting on a body equals the change in its kinetic energy".
I don't get it though, how does that help me solve this problem?

Doc Al

You want the work done, so see if you can figure out the change in the puck's KE. (You can't figure it out numerically, of course. Just symbolically.)

Initial KE of puck = ??
Final KE of puck = ??

tbyrne

ΔKE = KE final - KE initial = 1/2 mv2 final - 1/2 mv2 initial
Right? So, does this help me? What does that have to do with the hockey stick?

Doc Al

Sure. So what's the initial speed and KE of the puck?

The hockey stick is the object that did the work on the puck.

tbyrne

Well?? Initial speed is 0, I think, and KE is ?? I don't know, I am so confused. I am trying, but not getting it...

tbyrne

Does W=Delta E have anything to do with it?

Doc Al

Right. It starts out at rest.
How do you calculate KE? (You just gave the formula above.)

tbyrne

1/2 mv2 final-1/2 mv2 initial
But how do I do that without any numbers?

Doc Al

Using the definition of KE = 1/2mv², what's the initial KE? The initial velocity is 0, so what's the KE?

For the final velocity, what do they tell you the speed is? (Again, no numbers, just symbols.) So what's the final KE?

tbyrne

So, I guess initial KE would be 0 as well?
The speed is v. So, the final KE would be v-0??

Doc Al

Right! Since KE = 1/2m(speed)², the initial KE = 1/2m(0)² = 0
No. What's the final KE given the speed is just v. (It's much easier than you think.)

sgstudent

I thought initial speed should be the greatest since that's where the force is applied. Then once it stops then velocity is 0. Anyways the work done to do something will be equal to the energy change of an object be it KE, GPE or any forms of energy. Once it is used to work, then all that energy is 'channeled' to doing work. It is converted to work done by applied force so it covers all retardation such as friction or air resistance. Hence, that 1/2mv^2 is converted to work done by applied force to push the puck. However, the net work done is not equal to 1/2mv^2 due to the work done against friction.

Work done and energy are similar yet different. Similar in the sense that they have the same SI unit and magnitude. But different such that, at any point the puck has kinetic energy although lesser than at the original position. So using energy to explain the movement is like this: when the puck is first whacked, it has maximum kinetic energy. As it moves the kinetic energy of the puck reduces as energy is given off as heat and sound. Once all the energy is depleted, there is no more velocity, hence it stops moving. However, the work done explanation still holds true.

I hope this helps and if I'm wrong anywhere do point out my mistakes and correct them. Thanks for the help!

Doc Al

We are talking about the interaction of the stick with the puck. When the stick first makes contact, the initial speed of the puck is zero. It accelerates to its final speed due to the work done by the stick. Once the puck leaves the stick, it remains moving at that final speed. (There are no dissipative forces to worry about.)

The work done equals the change in KE.

sgstudent

But am I right in the sense if I take the whole system where the puck eventually stops? So in this case is it right to say that you are going into the micro interaction where I'm taking before the puck gets hit and at the point where the puck is bring hit? So the gain in kinetic energy is 1/2m ^2?

Is this question too simplified, or can this question come out in the o levels? Thanks for the help!

Doc Al

But you're changing the problem. (The original problem in this thread clearly asks for the work done by the stick when it hits the puck.) It sounds like you want to consider the puck starting from rest, getting hit by the stick, and then ending up at rest due to friction. In that case, what do you think the net work done is?