Aamc 9 ps #29

[coyotelime]

---

Members do not see this ad.

Question has a simply pulley with a block moving from point A to B.

"Which action involves more work: lifting a weight from A to B or lowering the weight from B to A?"

I understand that potential energy is mgh, and that you are lowering and raising by the same height. I had this problem in my physics lab actually too. However, the question does not tell you to neglect gravity. When you are pulling, gravity works against you and when you lower it, the weight of the block(mg), plays a role in facilitating the motion. What am I missing here? It's such a simple question but I missed it. Thanks!

 

[ThirdEye]

Question has a simply pulley with a block moving from point A to B.

"Which action involves more work: lifting a weight from A to B or lowering the weight from B to A?"

I understand that potential energy is mgh, and that you are lowering and raising by the same height. I had this problem in my physics lab actually too. However, the question does not tell you to neglect gravity. When you are pulling, gravity works against you and when you lower it, the weight of the block(mg), plays a role in facilitating the motion. What am I missing here? It's such a simple question but I missed it. Thanks!

W=Fd=mgh
Gravity doesn't change, mass doesn't change, and distance doesn't change. Only direction is changing.

In one instance, you are working against gravity. In the other instance gravity is doing the work.

 

[whiteshadodw]

W=Fd=mgh
Gravity doesn't change, mass doesn't change, and distance doesn't change. Only direction is changing.

In one instance, you are working against gravity. In the other instance gravity is doing the work.

i almost got this question wrong, but its almost as if they are asking for the MAGNITUDE of the work. if you consider just magnitude, then both are equal.

 

[johnmhh]

What confuses me about this question is: you're using the pulley to life the weight from A to B and in order to accomplish that, the tension force pulling on the object needs to be greater than the gravitational force or the weight of the mass at point A. So since this force is greater than the gravitational force, isn't the work done gonna be greater too?
I really need help with this.
Thanks in advance.

 

[thed0ctor]

I am stuck on this too.

If the tension force (upward) equals the weight of the object (downward), the object would not move at all because the forces cancel out. Therefore, I figured that the force to lift the object must be greater, thus making the work greater (Fd).

 

[milski]

The tension force needs to be greater to start it moving up but at some point it has to become smaller, so that the object stops at its final position. Calculating the exact work just based on forces and distances is actually quite hard. It's a lot easier to approach the problem from energetic point of view. Since the object is not moving neither in the initial or the final position, there is no change in kinetic energy and the work done on it by the tension force is just the change in potential energy, which does have the same magnitude, regardless of direction.

Keep in mind that the question (seems to be) asking about the work done by the tension force. If it was asking about the total work done on the weight, that would include the work done by gravity. According to work-kinetic energy theorem, the total work will be W_total=ΔKE=0.

 

[thed0ctor]

The tension force needs to be greater to start it moving up but at some point it has to become smaller, so that the object stops at its final position. Calculating the exact work just based on forces and distances is actually quite hard. It's a lot easier to approach the problem from energetic point of view. Since the object is not moving neither in the initial or the final position, there is no change in kinetic energy and the work done on it by the tension force is just the change in potential energy, which does have the same magnitude, regardless of direction.

Keep in mind that the question (seems to be) asking about the work done by the tension force. If it was asking about the total work done on the weight, that would include the work done by gravity. According to work-kinetic energy theorem, the total work will be W_total=ΔKE=0.

Yea, this only makes sense if I think about it as conservation of energy ( | Δ height | is same in both directions).

Thinking of it as a force being applied over a distance necessitates incorporating Fnet, not just Ftension. It is | Fnet | that is the same in both directions, not | Ftension |, but nobody has seemed to bring up Fnet or any sum of forces in their explanations or in the answer key... only "mg". 😕

 

[milski]

Yea, this only makes sense if I think about it as conservation of energy ( | Δ height | is same in both directions).

Thinking of it as a force being applied over a distance necessitates incorporating Fnet, not just Ftension. It is | Fnet | that is the same in both directions, not | Ftension |, but nobody has seemed to bring up Fnet or any sum of forces in their explanations or in the answer key... only "mg". 😕

Work does not depend on the way you calculate it - both approaches should give you the same result. Nobody has done it by using forces and distances, since it unnecessary complicates the calculation.

Consider a distance of h meters. Let's apply a F_t=m(h+g) for once second and then F_t=m(h-g) for another second. Then F_net=m(h+g)-mg=mh and a=mh/m=h. For one second, the displacement then is 1/2*at²=1/2h1²=h/2. For the next second F_net=m(h-g)-mg=m(h-2g), the displacement is another h/2, the final velocity is 0 and the work done is W=d₁F₁+d₂F₂=h/2*m(h+g)+h/2*m(h-g)=mgh. You can see that the additional work (compared to mgh) done during the acceleration is exactly equal to the work less than mgh done during the deceleration phase.

That's just one possible way of doing the acceleration/deceleration, you can use calculus the generalize it for any sort of motion which ends up with the same initial and final conditions. There is no need to make everything so complicated, since the energetics approach works much better here.