Problem: An object must gain 57 Joules of energy from a force of 3 Newtons being exerted on it parallel to its direction of motion. If the object does not dissipate any of the energy added to it, then over what distance must it be pushed?
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Solution: It is assumed that there is no dissipation of energy, so the work done by the 3 Newton force must be equal to the 57 Joules of energy used up. Since work equals force times distance, 57 Joules is equal to 3 Newtons multiplied by the distance. This could be written as an equation 57 J = ( 3 N)(dist). It is clear that dist = 57J/ 3N = 19 meters."
Generalized Response: Since the work `dW is equal to the product of the parallel force F and the displacement `ds, the force must be equal to the quotient F = `dW / `ds.
.
.
.
.
.
.
.
.
.
.
Figure description: The figure below shows the relationships among force, displacement and work for the case of force parallel to displacement. The relationship F = `dW / `ds tells us, for example, that if we wish to do a lot of work `dW over a short distance `ds we must exert a lot of force. The relationship `ds = `dW / F tells us, for example, that if we wish to do a lot of work `dW using a small force we had better be apparent to exert that force through a long distance. The relationship `dW = F `ds tells us, for example, that the more force we exert and the greater the distance over which we exert it, the more work we do.
