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# Work can be negative!

Work is a measurement of energy, so it may seem odd to think that a work can be negative — but it can!
Work is how much energy is done by a force over a distance. Suppose we needed to set up ice hockey goal nets. Initially the nets are standing still with zero velocity at the edge of the ice hockey rink. When you push a net, it will start moving at some speed; it has velocity. If there is a velocity then the net now has kinetic energy. Because the direction of the push and the movement of the net are in the same direction, there is now positive energy: the goalie net went from zero energy to some amount of positive kinetic energy. This is an example of positive work.
Suppose at the end of an ice hockey game an assistant helps put the goalie nets away by shoving the nets towards you. If you try to stop the net from moving, you will apply a force in the opposite direction that the net is moving. But the goalie net may be moving so fast that you are not able to stay still on the ice, and you move backwards. Your force is in the opposite direction as the movement. This is the opposite situation from the one above: the moving goalie net has kinetic energy, but loses kinetic energy as you slow it down. In this case, work is negative as the force applied to the goalie net is in the opposite direction of the net’s movement.
Remember when doing work calculations only the component of force parallel to the direction of movement is used in a work calculation.
For example, in this picture here as a person pushes the book across the table, neither the gravity force or the normal force do any work because there is no displacement in the y-direction, but the pushing force does do work because the pushing force is in the same direction as the movement of the book.

### EXAMPLE

Let look as some examples of calculating work all involving this crate of snakes weighing 50kg.
What is the work done by gravity as this crate of snakes is hoisted up 10 meters?
$W=Fd=\left(-mg\right)d=\left(-9.8\right)\left(50\right)10=-4900J$
This is negative work! Gravity does negative work here because gravity goes in the opposite direction as the displacement.
If the crate is moving upwards at a constant velocity, what is the work done by the tension in the rope as it is hoisted up 10 meters?
The phrase “constant velocity” always translates to zero acceleration. From drawing a free body force diagram, we can see the tension force is equal the gravity force, so T=mg.
$W=Fd=Td=\left(mg\right)\left(10\right)=4900J$
This is positive work. The tension force is in the same direction as the movement.
If the crate is moving upwards at a constant velocity, what is the net work done on the crate as it is hoisted up 10 meters?
$W=Fd=\left({F}_{g}+T\right)d=\left(-4900J+4900J\right)d=0$
The net work is zero because there is no increase or decrease in kinetic energy: the crate is moving at a constant velocity.

## Want to join the conversation?

• Isn't the unit for work supposed to be Joules instead of Newtons?
• I believe the notes are incorrect in a few other ways. The notes have "feet" instead of "meters." This "(mg)(10)=4900N" should read as "(mg)(10m)=4900J."
• I'm a little confused by the snake example. Isn't potential energy being put into the crate of snakes if it's higher than it initially started? Wouldn't this mean work is being done on it?
• There is work being done on the crate by both the force pulling it upwards and gravity, however the forces doing this work are equal and opposite, giving it a constant velocity and a net work of zero
• For the last example, I understand there is no change in kinetic energy but there is a change in gravitational potential energy so how can the net work be zero
• A bit confused: If Word= FD Cos(theta), then wouldn't Work at 90 degrees be 0?
• Can a system have positive work even if the applied force is not perfectly parallel with the direction of displacement?
• Shouldn't the last example say mg instead of Fg? If it is Fg, then it will be mg^2. Which doesn't make any sense?
(1 vote)
• This article is somewhat confusing, but the g in these equations is not a separate variable it is supposed to be a subscript. In other words, "Fg" is indicating the "force due to gravity".

Have a look at the figures earlier in the article and I think this will be clearer ...
(1 vote)
• So, if there's friction on the ground and I push a box along the ground at a constant velocity, I'm not doing any work on the box because the friction is pushing back with an equal force and thus no acceleration? (Trying to mirror the very last example given with the box being lifted at constant velocity)
(1 vote)
• From the last example if both works done by tension and gravity are equal to each other and they cancel out how is there still movement? If work is equal to F times distance d and there is still displacement going upwards, how is work zero? If it's moving upwards isn't tension greater than gravity?