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I'm having some trouble understanding the concept of negative work. For example, my book says that if I lower a box to the ground, the box does positive work on my hands and my hands do negative work on the box. So, if work occurs when a force causes displacement, how does negative work happen? Are my hands displacing anything?

Brandon Enright
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AndrewChang
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  • Displacement (and movement) is not always caused by the force that you want to find it's work. In your example of lowering a box, gravity must be considered. – Mostafa May 24 '13 at 22:21
  • Your book may be wrong. It's not going down that makes a negative work, but, as explained by joshphysics, it's the slowing down of the movement. – fffred May 25 '13 at 07:49
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    @ffred The book is right, since holding an object in a gravitational field results in a normal force from the hand, and since the motion is downward it is in fact negative work – yoel halb Apr 17 '14 at 16:19
  • Slightly tangential but it may help get a feeling for what is meant by negative work, have a look at the different conventions for writing out the first law of thermodynamics. You can see that what is defined as positive work under one convention is negative work in another. – ChrisM Feb 09 '15 at 01:42

2 Answers2

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In the context of classical mechanics as you describe, negative work is performed by a force on an object roughly whenever the motion of the object is in the opposite direction as the force. This "opposition" is what causes the negative sign in the work. Such a negative work indicates that the force is tending to slow the object down i.e. decrease its kinetic energy.

To be more mathematically precise, suppose that an object undergoes motion along a straight line (like in your example) under the influence of a force $\mathbf F$, then the work done on the object as it undergoes a small displacement $\Delta\mathbf x$ is $$ W = \mathbf F\cdot\Delta\mathbf x $$ where boldface means that the variable is a vector, and the dot represents dot product. From the definition of the dot product, we have $$ W = F\Delta x\cos\theta $$ Where $F$ is the magnitude of $\mathbf F$, $\Delta x$ is the magnitude of $\Delta \mathbf x$, and $\theta$ is the angle between $\mathbf F$ and $\Delta\mathbf x$. Note, in particular that the magnitudes are positive by definition, so the $\cos\theta$ is negative if and only of $\theta$ is between $90^\circ$ and $180^\circ$. When the angle has these ranges, the force has a component perpendicular to the direction of motion, and a component opposite to the direction of motion. The perpendicular component contributes nothing to the work, and the component opposite the motion contributes a negative amount to the work.

efx
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joshphysics
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  • Howdy. Physics noob here. Is it necessarily true that if θ is between 90∘ and 180∘ that there is a perpendicular component? Is the same true for Quadrant 1? I'm trying and failing to see it. – Ungeheuer Sep 19 '18 at 01:51
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Work is the component of force parallel to the direction of motion times the displacement. That component of force could of course point in the opposite direction of motion (anti-parallel). The work done by the force is positive in the first case and negative in the second. For instance, the direction of the force of gravity on a freely falling body (dropped from rest) points towards the center of the Earth which is also the direction of displacement while falling. Hence, the force of gravity is said to be do positive work on the falling body. The falling body also experiences a upwards drag force due to air resistance. Since the the drag force is being applied in the direction opposite to that of the motion, it's said to be doing negative work on the body.

David H
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