Calculating work

Calculating work

When you move an object, you do work. When you push an object very hard, but the object does not move, no work is done.

This is often confusing, and has to do with the meaning of "work" in physics. Work done is defined as the force on an object, times the distance that the object moves - in the direction of the force.

First, this means that work done refers to work "received" by the object. So if you push very hard, you do a lot of work on you muscles. If the object does not move, it does not receive any work from you. So we say no work is done.

Second, for work to be done, the object has to move "in the direction of" the force. If you push downwards on a trolley, while the trolley slides foward, then you are not doing any work. If movement is perpendicular to your force, then it is not "in the direction of" the force.

What if movement is at some angle to the force? Then part (a component) of the movement is perpendicular to the force, and part of it is parallel to (in the direction of) the force. We must only use the latter part.

How do we calculate this parallel component? We project the displacement of the object, onto a line along the force. This means we multiply the distance, by the cosine of the angle between movement and force.

Exercise. A force of 2 N pushes a book on a table. The book slides at a velocity of 0.5 m/s. It does not move any faster because of friction. What is the work done after 4 s? Answer. The distance moved is d = vt = 0.5 x 4 = 2m. So work done is W = Fd = 2 x 2 = 4 J. (In this case, work done does not produce more kinetic energy, since there is no change in velocity. It would produce 4 J of heat, because of the friction.)

Another common situation in which work is done is when a gas expands. This can be when air is wormed by the sun, or when steam is evaporated by boiling water. We can calculate the work done with the help of a simple example.

Consider a hollow cylinder, closed at one end, with a movable piston in the middle - like a syringe with the needle blocked. When the gas inside is heated, it expands and the piston moves. So the gas does work on the piston.

Suppose the pressure of the gas is p, the area of the piston is A, and the distance moved d. Then the force F = pA, and work done W = Fd = pAd. Ad is also the increase in volume ΔV of the gas. So we find the formula W = p ΔV.

This formula is also true for more complicated situations.

Exercise. When water boils, the steam pushes against an air pressure of 10⁵ Pa. If 0.1 m³ of steam is produced, how much work is done by the steam? Answer. We can use W = p ΔV = 10⁵ x 0.1 = 10⁴ J.

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