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Finding work using calculus

So far we have looked at the work done by a constant force. In the physical world, however, this often is not the case. Consider a mass moving back and forth on a spring. As the spring gets stretched or compressed it exerts more force on the mass. Thus the force exerted by the spring is dependent on the position of the particle. We will examine how to calculate work by a position dependent force, and then go on to give a complete proof of the Work-Energy theorem. Work Done by a Variable Force

Consider a force acting on an object over a certain distance that varies according to the displacement of the object. Let us call this force F(x) , as it is a function of x . Though this force is variable, we can break the interval over which it acts into very small intervals, in which the force can be approximated by a constant force. Let us break the force up into N intervals, each with length δx . Also let the force in each of those intervals be denoted by F 1, F 2,…F N . Thus the total work done by the force is given by: W = F 1 δx + F 2 δx + F 3 δx + ... + F N δx

Thus
W = F n δx
This sum is merely an approximation of the total work. Its degree of accuracy depends on how small the intervals δx are. The smaller they are, the more divisions of F arise, and the more accurate our calculation becomes. Thus to find an exact value, we find the limit of our sum as δx approaches zero. Clearly this sum becomes an integral, as this is one of the most common limits seen in calculus. If the particle travels from x o to x f then: F n δx = F(x)dx

Thus
W = F(x)dx
   

We have generated an integral equation that specifies the work done over a specific distance by a position dependent force. It must be noted that this equation only holds in the one dimensional case. In other words, this equation can only be used when the force is always parallel or antiparallel to the displacement of the particle. The integral is, in effect, quite...
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