The Student Room Group

Negative work

Just a simple question. Could anybody confirm that it is equivalent to treat this scenario in both of these ways: A person does work in lifting an arbitrary mass an arbitrary height. Is this equivalent to the gravitational field doing negative work on the mass. Similarly, when an electric (or gravitational) field does work on moving a charge/mass through a potential difference dv, is that equivalent to the charge (or mass) doing negative work on the field. So in mathematical notation, the charge Q moves through a potential difference of V, W=QV, where W is the work done by the field. The work done by the charge 'w' is therefore w=-QV?
(edited 9 years ago)
Original post by Protoxylic
Just a simple question. Could anybody confirm that it is equivalent to treat this scenario in both of these ways: A person does work in lifting an arbitrary mass an arbitrary height. Is this equivalent to the gravitational field doing negative work on the mass. Similarly, when an electric (or gravitational) field does work on moving a charge/mass through a potential difference dv, is that equivalent to the charge (or mass) doing negative work on the field. So in mathematical notation, the charge Q moves through a potential difference of V, W=QV, where W is the work done by the field. The work done by the charge 'w' is therefore w=-QV?


There's a thread here
http://www.thestudentroom.co.uk/showthread.php?p=51885483
which discussed the idea of negative work. (after a uncertain start!)

By convention in physics, work done is force x distance moved in the direction of the force.
If the displacement is in the opposite direction to the force the work done is negative.
eg. the braking force applied to a moving car.
The sign of the work then determines whether energy is transferred to or from the object doing the work. Negative work implies energy is transferred from, positive implies to.
Reply 2
Original post by Stonebridge
There's a thread here
http://www.thestudentroom.co.uk/showthread.php?p=51885483
which discussed the idea of negative work. (after a uncertain start!)

By convention in physics, work done is force x distance moved in the direction of the force.
If the displacement is in the opposite direction to the force the work done is negative.
eg. the braking force applied to a moving car.
The sign of the work then determines whether energy is transferred to or from the object doing the work. Negative work implies energy is transferred from, positive implies to.


Ah, didn't see this, sorry. I knew there was a concept of negative work, I just wanted to see if it was correct in saying that the charge does negative work on the field if the field does work on the charge. So basically a consequence of Newton's third law?
Original post by Protoxylic
Ah, didn't see this, sorry. I knew there was a concept of negative work, I just wanted to see if it was correct in saying that the charge does negative work on the field if the field does work on the charge. So basically a consequence of Newton's third law?


In the case of fields (grav or elec) the work (positive) is done by the field on the mass or charge when it moves it in the direction of the force from the field. (Either attractive or repulsive.)

So, taking gravitation, if I drop an object and allow it to fall freely in the field, the field is doing positive work on the object and the field is transferring energy to the object. It's giving it kinetic energy.
If I throw the object upwards in the field, the field does negative work on the mass (decelerating force in opposite direction to motion) and this transfers energy from the mass (takes away its kinetic energy) and stores it in the field as potential energy.
It's really just a convention in physics to make sure the book-keeping is in order and we can account for where the energy is going and make sure it's conserved.
In most practical examples you meet, it probably doesn't matter if the work is positive or negative. It should be clear from the context what is going on. The maths (positive or negative sign - as I stated in that other thread) will take care of it if you do need to think about positive or negative work.
Reply 4
Original post by Stonebridge
In the case of fields (grav or elec) the work (positive) is done by the field on the mass or charge when it moves it in the direction of the force from the field. (Either attractive or repulsive.)

So, taking gravitation, if I drop an object and allow it to fall freely in the field, the field is doing positive work on the object and the field is transferring energy to the object. It's giving it kinetic energy.
If I throw the object upwards in the field, the field does negative work on the mass (decelerating force in opposite direction to motion) and this transfers energy from the mass (takes away its kinetic energy) and stores it in the field as potential energy.
It's really just a convention in physics to make sure the book-keeping is in order and we can account for where the energy is going and make sure it's conserved.
In most practical examples you meet, it probably doesn't matter if the work is positive or negative. It should be clear from the context what is going on. The maths (positive or negative sign - as I stated in that other thread) will take care of it if you do need to think about positive or negative work.


Right, I see, cheers.
Reply 5
Original post by Stonebridge
In the case of fields (grav or elec) the work (positive) is done by the field on the mass or charge when it moves it in the direction of the force from the field. (Either attractive or repulsive.)

So, taking gravitation, if I drop an object and allow it to fall freely in the field, the field is doing positive work on the object and the field is transferring energy to the object. It's giving it kinetic energy.
If I throw the object upwards in the field, the field does negative work on the mass (decelerating force in opposite direction to motion) and this transfers energy from the mass (takes away its kinetic energy) and stores it in the field as potential energy.
It's really just a convention in physics to make sure the book-keeping is in order and we can account for where the energy is going and make sure it's conserved.
In most practical examples you meet, it probably doesn't matter if the work is positive or negative. It should be clear from the context what is going on. The maths (positive or negative sign - as I stated in that other thread) will take care of it if you do need to think about positive or negative work.


Just to clarify. If you use the equation delta W=mdeltaV. That is the work done by you. Say if you drop a ball in a gravitational field, the change in potential will be negative, so the work done by yourself is negative.
(edited 9 years ago)
Original post by Protoxylic
Just to clarify. If you use the equation delta W=mdeltaV. That is the work done by you. Say if you drop a ball in a gravitational field, the change in potential will be negative, so the work done by yourself is negative.


You don't do any work if you drop the ball. The field does the work.
The energy is transferred from the field to the ball. So the ke of the ball increases and the energy has come from the field. This is expressed as a loss in potential energy. Potential energy being energy stored in the field.
If you lift the ball up, then you do work. The force you apply is in the direction of motion so you do positive work and potential energy is stored in the field as a result. Energy is transferred from you to the field.
You have to be careful that you identify who or what is applying the force and in which direction. Then look at where the energy is being transferred to and from.
(edited 9 years ago)
Reply 7
Original post by Stonebridge
You don't do any work if you drop the ball. The field does the work.
The energy is transferred from the field to the ball. So the ke of the ball increases and the energy has come from the field. This is expressed as a loss in potential energy. Potential energy being energy stored in the field.
If you lift the ball up, then you do work. The force you apply is in the direction of motion so you do positive work and potential energy is stored in the field as a result. Energy is transferred from you to the field.
You have to be careful that you identify who or what is applying the force and in which direction. Then look at where the energy is being transferred to and from.


So the equation deltaW=mdeltaV in the case of dropping a ball in a gravitational field the deltaW term will be negative because the change in potential is negative. So W wouldn't be work in this case, it would be potential energy?
Reply 8
Original post by Protoxylic
So the equation deltaW=mdeltaV in the case of dropping a ball in a gravitational field the deltaW term will be negative because the change in potential is negative. So W wouldn't be work in this case, it would be potential energy?


That deltaW is describing the work done by the field for a change in potential. If your change in potential is negative, so will your change in work. You're linking the work done by a human dropping a ball to the fields work wrongly I assume, the person is in no contact with the ball, you're purely describing the fields action upon the ball, in the case that it's free falling
Original post by Protoxylic
So the equation deltaW=mdeltaV in the case of dropping a ball in a gravitational field the deltaW term will be negative because the change in potential is negative. So W wouldn't be work in this case, it would be potential energy?


You have to be sure which "work" you are talking about. Work done by you or work done by the field.
Potential, here, is defined as work done on a unit mass getting it to a point in the field. It's work done by something else, not the work done by the field. This is why potential is negative in a gravitational field. Starting from the fact that it's defined as zero at infinity, you have to do positive work on a mass to get it to infinity.
The dW in that equation, if you define potential as I have stated, is not the work done by the field. If you were to raise the mass then dW would be positive as you would need to do work on it.

Quick Reply

Latest