concept question on work

[thebillsfan]

Are you only doing work on an object if you increase that object's energy?

I have an example. The first paragraph explains why it would not be a situation of doing work on a box, and the second would be why it WOULD. Someone explain the difference btw the two, please.

If you hold a box above your head at the same height and walk in a straight line, youre doing no work on the box bc it's at the same height (same potential energy) and when you stop walking the box is not moving so there's no kinetic energy added, either.

this makes sense to me...BUT, there's one issue. I also know that work is force*d, where force is in the direction the box traveled. If youre walking around with a box above your head, you obviously HAVE to be applying a force to the box in the direction that youre moving. so you would be doing work on the box.

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[thebillsfan]

Also, how does this relate to the work energy theorem...can't you do work on an object without increasing its kinetic energy?

 

[the freest man]

if you have a box sitting on a perfectly horizontal table, and you push it with X force for 10cm, then W = 10X, but since it is at the same "h", the potential energy will not have changed. Also, when the box starts and stops it has v = 0, making the end KE = 0 too.

Correct me if I am wrong here.

 

[inaccensa]

Are you only doing work on an object if you increase that object's energy?

I have an example. The first paragraph explains why it would not be a situation of doing work on a box, and the second would be why it WOULD. Someone explain the difference btw the two, please.

If you hold a box above your head at the same height and walk in a straight line, youre doing no work on the box bc it's at the same height (same potential energy) and when you stop walking the box is not moving so there's no kinetic energy added, either.

this makes sense to me...BUT, there's one issue. I also know that work is force*d, where force is in the direction the box traveled. If youre walking around with a box above your head, you obviously HAVE to be applying a force to the box in the direction that youre moving. so you would be doing work on the box.

If you think this interms of W =Fd costheta. If there is no displacement of the object, there is no work done. If the force is applied perpendicular there is no work. In case of walking around with a box on your head. There is no net displacement. Thus there is no work. I really don't think this in terms of energy. Hope it helps

 

[inaccensa]

if you have a box sitting on a perfectly horizontal table, and you push it with X force for 10cm, then W = 10X, but since it is at the same "h", the potential energy will not have changed. Also, when the box starts and stops it has v = 0, making the end KE = 0 too.

Correct me if I am wrong here.

I'm confused too. I that is the case, PE is converted to KE. However, since the object is still at a height h, wouldn't the PE be constant throughout the displacement? But KE must be coming from somewhere!!

 

[inaccensa]

Also, how does this relate to the work energy theorem...can't you do work on an object without increasing its kinetic energy?

Raise an object from height 5 to height 10. The PE will increase, but the work done will not be zero.

 

[the freest man]

KE for the box is from the F moving it, but when the force is done moving the box, there is a v = 0, and thus KE = 0

 

[inaccensa]

KE for the box is from the F moving it, but when the force is done moving the box, there is a v = 0, and thus KE = 0

Yes I understand that part. But I'm trying to understand what happens during the whole displacement. I know that PE must be converted to KE, but then shouldn't the PE decrease and how can that happen when the object is at the same height

 

[thebillsfan]

KE for the box is from the F moving it, but when the force is done moving the box, there is a v = 0, and thus KE = 0

the KE is zero because of FRICTION but you still did work on the box (even though that work was taken away by friction).

but if you raise a box from 5cm to 10cm the work is not zero, is it?

 

[thebillsfan]

If you think this interms of W =Fd costheta. If there is no displacement of the object, there is no work done. If the force is applied perpendicular there is no work. In case of walking around with a box on your head. There is no net displacement. Thus there is no work. I really don't think this in terms of energy. Hope it helps

not walking around in a circle---walking in a line so that there IS a net displacement

 

[inaccensa]

the KE is zero because of FRICTION but you still did work on the box (even though that work was taken away by friction).

but if you raise a box from 5cm to 10cm the work is not zero, is it?

Work = change in PE, so it should not be zero. I don't know what I was posting..sorry

 

[inaccensa]

not walking around in a circle---walking in a line so that there IS a net displacement

but its u who's moving, not the box. The box is technically at the same height rt.

 

[the freest man]

No, the work would be >0. Lifting a box from underneath will displace it vertically and do work. Carrying a box around in the air does not do work, as the force (parallel to the floor) is at 90* to the box.

Pushing a box on a table would still generate KE though. But, I believe it is all lost as heat due to friction, yes.

 

[thebillsfan]

that's what i'm saying though...the force can't all by 90 degrees to the box. there's some component of force on your hands that's pushing the bo in whatever direction youre walking

 

[inaccensa]

that's what i'm saying though...the force can't all by 90 degrees to the box. there's some component of force on your hands that's pushing the bo in whatever direction youre walking

but wouldn't that cancel. The weight mg = force upward

 

[inaccensa]

can some one please explain the changes that are occuring when the box is displaced along a straight line at a height h.Energy is really confusing

 

[wanderer]

Are you only doing work on an object if you increase that object's energy?

I have an example. The first paragraph explains why it would not be a situation of doing work on a box, and the second would be why it WOULD. Someone explain the difference btw the two, please.

If you hold a box above your head at the same height and walk in a straight line, youre doing no work on the box bc it's at the same height (same potential energy) and when you stop walking the box is not moving so there's no kinetic energy added, either.

this makes sense to me...BUT, there's one issue. I also know that work is force*d, where force is in the direction the box traveled. If youre walking around with a box above your head, you obviously HAVE to be applying a force to the box in the direction that youre moving. so you would be doing work on the box.

To me work is anything that changes the (potential or kinetic) energy in a system. So if you are lifting a box you are raising its potential energy and doing work. When you drop the box it accelerates and turns its potential energy into kinetic energy so gravity is doing work on the box. When something slows down due to friction, the friction is doing work on the object as well. When you move a box horizontally such as when you are pushing it across a frictionless surface and it has zero velocity at the end zero net work was done because its potential and kinetic energy are unchanged. This may seem counter intuitive because the box does not move by itself, but in reality an infinitesimally small push (which can be assumed to have a force of about zero) will move the object but will not change its kinetic or potential energy. In a more realistic example if you are moving a box (while picking it up) and it has zero kinetic energy at the end it is because you did positive work in accelerating the object initially and did negative work at the end so that the net work is zero.

 

[kentavr]

There are several details in this question.
1. If you start moving the box from speed=0 to some speed V, then it get kinetic energy mv^2/2 and friction force(through you) made this work.
2. When you reach speed V and keep it stable, no work is done. Since in ideal world the body suppose to keep its constant speed with no force.
3. The normal reaction force(support) is perpendicular to your movement and cos(90) = 0.
4. The paradox is that you still get tired. But that is because you not keeping the box at constant height. Micro muscular movements and changes in your gravity center during walking will produce the work in the vertical movements multiplied your weight and box's weight. (Approx)
HTH.