Zero work when carrying object?

[sickle]

I never seemed to wrapped my head around the concept of zero work when carrying an object horizontally.

I understand that work is a transfer of energy to an object when it is displaced by a force.

Why is the work zero when you say carry a sandbag horizontally? Yes, your lifting force (counteracting g) is perpendicular, but isn't there also a horizontal force opposing friction?

So is pushing a an object at a constant velocity horizontally also 0 work?

From the textbook, it seems that work is dependent the net force acting on the object, not the force applied by you correct?

 

[disregardthat]

It follows from the definition of work. The force in the direction the sandbag goes is 0, so the force you apply to the sandbag is only vertical, but it doesn't move vertically so the work is 0. Work is only done if the force actually moves the object in its direction.

 

[diazona]

I never seemed to wrapped my head around the concept of zero work when carrying an object horizontally.

I understand that work is a transfer of energy to an object when it is displaced by a force.

Why is the work zero when you say carry a sandbag horizontally? Yes, your lifting force (counteracting g) is perpendicular, but isn't there also a horizontal force opposing friction?

So is pushing a an object at a constant velocity horizontally also 0 work?

From the textbook, it seems that work is dependent the net force acting on the object, not the force applied by you correct?

Basically yes. Here's one way to think about it: the sandbag (or other object) does not gain any energy, thus no net work is done on it.

However, there is also a concept of work done by a particular force. This is not talking about the net work, but about an amount of work that can be associated with one individual force. If the object does not gain any energy (i.e. its speed and height do not change), the net work done on it is zero, but the work done on it by each individual force is not necessarily zero. If the net work is zero, that simply means that the amounts of work done by all other forces must cancel out. (Forces acting against the object's motion do negative work)

In the case of the sandbag, if there is friction, then you will have to apply a force to keep the bag moving at constant velocity. And you do indeed do work on the bag through that force. However, friction does the exact opposite amount of work (which should make sense because it's the same amount of force, in the opposite direction), so the net work done on the bag is zero, and thus it doesn't change energy.

Mathematically, you can think of it like this: there is an amount of work associated with each individual force that acts on an object, and that amount can be calculated from the formula
W = \int \vec{F}\cdot\mathrm{d}\vec{x}
(this tells you work done on the object)
The same kind of formula also applies to the net force:
W_\text{net} = \int \vec{F}_\text{net}\cdot\mathrm{d}\vec{x}
Note that just as the net force is the (vector) sum of all the individual forces, the net work is the (scalar) sum of the work done by all those individual forces:
W_\text{net} = \sum W
It is this net work that changes the kinetic energy of the object,
W_\text{net} = \Delta K = \Delta\biggl(\frac{1}{2}mv^2\biggr)

 

[sickle]

Basically yes. Here's one way to think about it: the sandbag (or other object) does not gain any energy, thus no net work is done on it.

However, there is also a concept of work done by a particular force. This is not talking about the net work, but about an amount of work that can be associated with one individual force. If the object does not gain any energy (i.e. its speed and height do not change), the net work done on it is zero, but the work done on it by each individual force is not necessarily zero. If the net work is zero, that simply means that the amounts of work done by all other forces must cancel out. (Forces acting against the object's motion do negative work)

In the case of the sandbag, if there is friction, then you will have to apply a force to keep the bag moving at constant velocity. And you do indeed do work on the bag through that force. However, friction does the exact opposite amount of work (which should make sense because it's the same amount of force, in the opposite direction), so the net work done on the bag is zero, and thus it doesn't change energy.

Mathematically, you can think of it like this: there is an amount of work associated with each individual force that acts on an object, and that amount can be calculated from the formula
W = \int \vec{F}\cdot\mathrm{d}\vec{x}
(this tells you work done on the object)
The same kind of formula also applies to the net force:
W_\text{net} = \int \vec{F}_\text{net}\cdot\mathrm{d}\vec{x}
Note that just as the net force is the (vector) sum of all the individual forces, the net work is the (scalar) sum of the work done by all those individual forces:
W_\text{net} = \sum W
It is this net work that changes the kinetic energy of the object,
W_\text{net} = \Delta K = \Delta\biggl(\frac{1}{2}mv^2\biggr)

k that clears it up a lot! thanks!
our crappy textbook never actually defines individual work done on object by a force and net work. this always made the lifting objects vertically questions so confusing!

 

[sickle]

so on this note (and to clear up things once and for all):

1) if the object has zero displacement, all work (net and individual) are all zero

2) if displacement is nonzero and velocity is constant: net work is zero because net force is zero (but individual work can be non zero e.g. applied, gravity)

3) if displacement is nonzero and acceleration is non-zero: net work is non zero

Thus, in the lifting objects vertically at constant velocity questions, net work is 0 right? (and the questions always asks for work you did or gravity did)

Applying the same logic, carrying/pushing an object at constant velocity is also 0 net work but non zero individual work (by applied and friction) right?

So what exactly is work? It is energy transferred in general or energy transferred to the object? (does the two above examples differ because the vertical example is creating gravitation potential while the horizontal is creating heat)?

 

[D H]

so on this note (and to clear up things once and for all):

1) if the object has zero displacement, all work (net and individual) are all zero

2) if displacement is nonzero and velocity is constant: net work is zero because net force is zero (but individual work can be non zero e.g. applied, gravity)

3) if displacement is nonzero and acceleration is non-zero: net work is non zero

1) Correct. No motion = no work, by definition.

2) Incorrect. Raising an object vertically at a constant velocity requires work.

3) Incorrect. A pair of objects gravitationally orbiting one another circularly do zero work on one another.

To have non-zero work the force has to have a non-zero component in the direction of the velocity vector. In my counterexample for your second statement, a force is needed to counteract gravity. That force is parallel to the velocity vector and hence is performing work. For an object in a uniform circular orbit the force vector is always perpendicular to the velocity vector. So even though there is acceleration in this case, there is no work.

 

[diazona]

1) if the object has zero displacement, all work (net and individual) are all zero

Yes, exactly.

2) if displacement is nonzero and velocity is constant: net work is zero because net force is zero (but individual work can be non zero e.g. applied, gravity)

Yes. (It's a little unusual to find gravity doing work in a situation when velocity is constant, though it's certainly possible. Friction is a more typical example of a force that does work when the net work is zero.)

3) if displacement is nonzero and acceleration is non-zero: net work is non zero

Almost. It is possible that you have a nonzero acceleration which is perpendicular to the velocity, and in that case, the direction of the velocity is changing but its magnitude stays constant. The net work done would be zero, because work is related to the change in speed (i.e. magnitude of velocity).

In any other case where there is a nonzero acceleration, the speed would be changing and the net work would be nonzero.

Thus, in the lifting objects vertically at constant velocity questions, net work is 0 right? (and the questions always asks for work you did or gravity did)

Right. Of course, asking what the net work is in such a case is a perfectly valid test question, just to see if you know what you're talking about. (The answer would be 0 of course)

Applying the same logic, carrying/pushing an object at constant velocity is also 0 net work but non zero individual work (by applied and friction) right?

Right. But it's possible that there really are no forces at all (e.g. an object flying through empty space), and in that case there is no individual work done either.

So what exactly is work? It is energy transferred in general or energy transferred to the object? (does the two above examples differ because the vertical example is creating gravitation potential while the horizontal is creating heat)?

The work done on a system is energy transferred to that system by mechanical means. And the work done by a system is energy transferred from that system by mechanical means. Here "mechanical" means just a push or a pull - basically, a force. So you could think of work as energy that's transferred to the system through a force. It's defined that way to distinguish it from heat, which is energy transferred by thermal contact. That's the other way in which energy can be transferred to or from a system. If a system gains or loses heat, that doesn't change its overall motion at all.

Note that energy can leave one system as work and arrive in another system as heat. This is exactly what friction does: it takes mechanical energy from the moving object and transfers it to the surroundings in the form of heat. But it doesn't go the other way around: you can't turn the heat back into work. You can't get the object back to the speed it started at without adding additional energy into the system. (This is a consequence of the 2nd law of thermodynamics, in case you were curious)

In your gravitational example, on the other hand, you can get all the energy back. If you change an object's height, its speed will change, but you can restore the object to its original speed by letting it return to its original height. You don't have to put any extra energy back into the system. That's probably the main difference between your two examples.

 

[diazona]

To have non-zero work the force has to have a non-zero component in the direction of the velocity vector. In my counterexample for your second statement, a force is needed to counteract gravity. That force is parallel to the velocity vector and hence is performing work.

But gravity is performing the opposite amount of work, since it is a force of equal magnitude acting in the opposite direction. So in the language of this discussion, the net work (work done by gravity + work done by force counteracting gravity) is zero. The work performed by the force counteracting gravity is what we're calling "individual work," which sickle did correctly say could be (and would have to be) nonzero.

 

[sickle]

1) Correct. No motion = no work, by definition.

2) Incorrect. Raising an object vertically at a constant velocity requires work.

3) Incorrect. A pair of objects gravitationally orbiting one another circularly do zero work on one another.

To have non-zero work the force has to have a non-zero component in the direction of the velocity vector. In my counterexample for your second statement, a force is needed to counteract gravity. That force is parallel to the velocity vector and hence is performing work. For an object in a uniform circular orbit the force vector is always perpendicular to the velocity vector. So even though there is acceleration in this case, there is no work.

3) you are correct, i forgot about centripetal motion.
2) in example two, i meant total work on object (not applied work by person). because the object has no acceleration and thus no net force, there is also no net work.
Individual work however i think is nonzero (such as the work needed by person to counteract gravity). Is this correct?

 

[sickle]

I think i finally understand my horrible textbook! i think ...

The textbook always mixes up work by a single force (e.g. applied) and net without any warning whatsoever.

In the example of work done to an object by raising it vertically a constant speed, it implies individual work (effort of your hand).
In most horizontal problems however, it implies net work (thus we use the net force = applied - frictional).

Furthermore, the textbook ignores air resistance so it assumes the objects carried in your hand are frictionless (thus no force is required to to keep moving and no individual work). However, objects you push in direct contact with the ground suffer kinetic surface friction which the book does not ignore. Therefore you need a force (and therefore has a non zero individual work).

Note that all cases above actually have 0 total (net) work!

 

[D H]

But gravity is performing the opposite amount of work ...

Depends on what you mean by work. Is it change in kinetic energy (work energy theorem) or change in mechanical energy (work by nonconservative forces)? The answer is yes. Both meanings are in use.