# I have trouble understanding work

I am just starting out in physics. I study in germany, so excuse me if I get some terms wrong.

I am trying to understand why 'Work = Force * Way' and I think I just have some trouble imagining what work is. I have learned that force is that which can accelerate a mass, which excludes things like friction or gravity, so I just imagine it like "in space you need 1 N for 2 seconds to accelerate a 1kg mass to about 2m/s". This seems like a logical way to measure Force because things like inertia and acceleration are universal concepts.

But Force * Way? Does this still exclude friction and gravity? If some objects are accelerated by a certain force along a path, some will take longer and some will go faster, does that mean that the objects that take longer experience more of the work than the fast ones? Why not Force * Time, like in my mass-in-space example.

In all the examples I ever hear, people talk about lifting a mass from the ground and defining that as the work. So is work the force needed to 'push' an object against another (counter-directional) force for a certain distance? Is that 'other' force always gravity?

(Same goes for potential energy: A lot of examples of potential energy use gravity as some universal part of what potential energy is (?)..)

One thing that may help you is to understand that work is an expenditure of energy. For instance, if you push a rock up a hill, you exerted a force over a distance, so you had to use energy. But the coat hanger in your closet isn't expending energy, even though it's exerting a force to hold itself up over a period of time - because it isn't moving over any distance. (If you were to hold yourself up on the coat rack using your arms, you would still use energy, but that's because of biology - your muscles are always moving a little when they're exerting force, even when you're just trying to hold your body in place.)

• Thanks for the reply. Could it be said that the F in the work definition is the force that is 'working against me'? Nov 3 '15 at 2:55
• That depends whether you're talking about the work done by you or the work done on you. Nov 3 '15 at 3:05
• I was thinking work done by me Nov 3 '15 at 3:10
• Then you want to consider the force that you're exerting. Of course, when you exert a force on something, it exerts a force back on you (Newton's 3rd law), but the direction of the force matters here, because we're taking the dot product of the force with the displacement. So for instance if the force is in the same direction as the movement then you do positive work, whereas if the force is in the opposite direction of the movement then you do negative work. Nov 3 '15 at 17:03
• For example, if you're behind a cart pushing it then you are giving energy to the cart, but if you stand in front of a cart as it moves towards you and push against it then you are taking energy away from the cart. Nov 3 '15 at 17:08

First of all, forces accelerate an object when the net force is not zero. If friction is present, it actually does accelerate an object. Acceleration is defined as the change of velocity over time. This is not limited to increasing speed. Gravity also can accelerate an object.

Work is just defined as $Fd$ (force times distance) just because it is useful. Work could be called force*time but that is called impulse. Naming is arbitrary. But each have different properties.

Regarding including gravity and friction, using those forces in the work equation is perfectly fine and can be useful in many cases

• Thanks for the response. I am not really sure I understand, though. So is work just 'virtual' (i.e. not actually something to be measured)? Nov 3 '15 at 2:12
• @JohnSmith no work can be measured. But I was saying that force*distance just happens to be called work. It can be called something else as well. To measure work, you measure the force magnitude and multiply by the distance Nov 3 '15 at 2:16
• Can pure work be imagined as something, though? Like in the mass-in-space example, or was that example also not correct? Nov 3 '15 at 2:22
• @JohnSmith I don't understand. Your math is correct. Actually you have discovered the impulse-momentum theory. Nov 3 '15 at 2:26