Help, Calculating work done and what if you have too little or too much?

Question

Hey. My experiment is calculating the work done on different materials, I have obtained different results for the work done for each material. However the problem is, what is the ideal amount of work done in certain senario's? Is there a problem with too much or too little work done, which is my real question. Any help seriously appreciated.

Answer 1

Presuming you are talking about the work equation J = Fd; where J is work in Joules, F is force in Newtons, and d is distance moved in meters, then the answer is...yes you can do too much work or not enough. Why?

Because J/F = d; which means you work to move a body a distance d. If you were to push a mass with a fixed force F, but put in less work (J) than necessary, your desired distance moved would not be achieved. Conversely, if you work more than required, you'd overshoot your desired distance. Note, too much or too little work requires that a given d is specified as a required distance to move the mass.

Because you failed to describe your experiment in detail, I can only presume you are pushing various masses of various material along some surface for some specified distance (d). Then you are calculating the amount of work done J = Fd for each mass of different material. F in these experiments would be sliding friction force = kN; where k is the coefficient of sliding friction and N is the weight of each mass perpendicular to the surface.

What you should discover is that the F's for each material will be different; so that, even if your d is the same for each trial, work will vary with the force exerted to slide the body. That is, more F means more work and vice versa. In each case, the amount of work performed is just sufficient to move the mass d meters...no more, no less. But if you push too far, you will have put in too much work...too short, not enough work.

The F's will differ because of different k's and different N's. In this experiment, if d is not specified, then the work done in each case is just sufficient (no more or less) than what was needed to move each different d distance.