Talk:Physics Study Guide/Work

''When work is applied to an object or a system it adds or removes kinetic energy to or from that object or system. More precisely, a net force in one direction, when applied to an object moving opposite or in the same direction as the force, kinetic energy will be added or removed to or from that object. (Someone please explain this to me though because energy is not a vector.) Note that work and energy are measured in the same unit, the Joule (J).''
 * That's why you have to use a scaler product! Work is the scaler product of force and displacement. Theresa knott 09:37, 10 Nov 2003 (UTC)

Please, what is a scalar product ? If I dont know this, many other people wont know this, and we could use a good, concise definition here on the site. Thanks, --Karl Wick

OK I'll have a go. Let's take gravity as our force. If you jump out of an aeroplane and fall you will pick up speed. (for simplicity sake let's ignore air drag). To work out the kinetic energy at any point you just multiply the value of the force of gravity, by the distance moved. So for example a 100kg man falling a distance of 10m will have 1000kJ extra kinetic energy. We say that the man has had 1000kJ of work done on him by the force of gravity.

Notice that energy is not a vector, it has a value but no direction. Gravity and dispacement are vectors. They have a value plus a direction.(In this case their directions are down and down respectively) The reason we can get a scalar energy from vectors gravity and dispacement is because in this case they happen to point in the same direction. Gravity acts downwards and displacement is also downwards.

When two vectors point in the same direction, you can get the scalar product by just multiplying the value of the two vectors together and ignoring the direction.

What happens if they don't point in the same direction?

Consider a man walking up a hill. Obvoiusly it takes energy to do this because you are going against the force of gravity. The steeper the hill the more energy it takes every step, this is something we all know unless we live on a salt lake.

In a situation like this we can still work out the work done. In the diagram the green lines represents the displacement. To fond out how much work against gravity the man does we work out the projection of the displacement along the line of action of the force of gravity. In this case it's just the y component of the man's displacement. This is where the cos&theta; comes in.

If the two forces do not point in the same direction, you can still get the scaler product by multiplying the projection of one force in the direction of the other force. A(Bcos&theta;)

I hopes this makes sense. Theresa knott 17:16, 10 Nov 2003 (UTC)

But don't these situations only apply in the specific instances that you have described? Surely, the relationship might not always be cos theta, rather sin theta. Sorry if I'm mistaken. Also, the formula describing work done = deltaKE + deltaPE is also a specific instance, where there are no losses, for example air resisitance or friction. Shouldn't this be mentioned? If you're going into the preciseness of vectors and scalars or whatever, shouldn't this be mentioned? Finally, I'd just like to say it's a great job you're doing here! (I don't have an account on Wikibooks) --81.157.105.216 18:53, 9 Jan 2005 (UTC)