Kinetic Energy Change of a Cart on a Frictionless Surface

[DavidAp]

There's a cord attached to a cart that can slide along a frictionless horizontal rail aligned along an x axis. The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height h = 1.3 m, so the cart slides from x1 = 5.0 m to x2 = 1.0 m. During the move, the tension in the cord is a constant 29.0 N. What is the change in the kinetic energy of the cart during the move?

Answer 102 J

I have no idea how to approach this problem. I know the equation W = ΔK + ΔU and since we are looking for ΔK I was thinking that I can rearrange the equation to look like,
W - Uf + Ui = ΔK.

However, this raises the question: what's the Work done in this equation? What is the potential energy? From these question I thought I would solve for Work using W = ∫Fdr so,
W = 29N(4m) = 116 Nm

However, something about the way I solved for Work doesn't feel right. I'm still relatively new to Work and am having difficulty putting two and two together when it comes to this topic. If someone can guide me on this problem and, more importantly, explain to me why you chose the equations you did I would greatly appreciate it!

Thank you for taking the time to review my question.

 

[cepheid]

Something seems like it's missing in this problem. I don't understand the phrase, "The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height h = 1.3 m". The last part doesn't seem like a complete thought to me.

By the way, it's NOT true that W = ΔK + ΔU.

W = ΔK. This is the work-energy theorem. Now, if the force that acts is conservative, then conservation of energy says that ΔK = -ΔU, and hence:

W = -ΔU.

 

[DavidAp]

Something seems like it's missing in this problem. I don't understand the phrase, "The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height h = 1.3 m". The last part doesn't seem like a complete thought to me.

By the way, it's NOT true that W = ΔK + ΔU.

W = ΔK. This is the work-energy theorem. Now, if the force that acts is conservative, then conservation of energy says that ΔK = -ΔU, and hence:

W = -ΔU.

I double checked the text and I didn't forget anything, that's how they worded it. Here's a picture of a diagram they provided though!

So ΔK = ΔU but I still don't know how to calculate PE :/.

 

[cepheid]

I double checked the text and I didn't forget anything, that's how they worded it. Here's a picture of a diagram they provided though!

Yes, but the diagram was everything here though! Before, the wording of the problem made no sense. I though the cord was horizontal, and I had no idea what "cord height" referred to. Now I do.

So ΔK = ΔU but I still don't know how to calculate PE :/.

NO. That's not what I said. I said ΔK = -ΔU, for a conservative force like gravity. But who says you have to calculate PE at all for this problem? It is totally irrelevant! We're not dealing with gravity, or any other conservative force, and the gravitational PE of the block does not change during the motion. I'm not sure why PE was even brought up.

Just use the work-energy theorem. If you can figure out how much work was done, you'll know the change in kinetic energy. So how do you figure out how much work was done? Well, since x decreases to the left, it seems pretty clear that x-positions are measured from the "cliff" i.e. the origin of the coordinate system (x=0, y=0) is at the base of the cliff. This means that the cord rises 1.3 m vertically over a horizontal distance of [STRIKE]6.00 m[/STRIKE] 5.00 m. This information will allow you to figure out the angle of the rope above the horizontal.

Just made a small EDIT above.

 

[cepheid]

Aha! I just realized that this problem is a lot more tricky than I first realized, because the rope angle that I mentioned in my previous post will not be constant, but will change with x.