# Work and a Spring

1. Oct 31, 2006

### mbrmbrg

In the figure below, we must apply a force of magnitude 89 N to hold the block stationary at x = -2.0 cm. From that position we then slowly move the block so that our force does +3.3 J of work on the spring-block system; the block is then again stationary. What is the block's position? (There are two answers.)

The figure is a block resting on a horizontal surface, attatched to a horizontal spring.

-our force=spring force=-kd
so -89J=-k(-0.02m)==>k=-4450N/m

I then said that $$W=-(k\frac{x_f^2}{2}-k\frac{x_i^2}{2})$$, subbed in W=3.3 J, k=-4450N/m, x_i=-0.02m, and got the right answer (which is +/- 4.3cm).
But why did that equation work? 3.3 J is our work, not the spring's?

2. Oct 31, 2006

### PhanthomJay

Yes, that is the work done by the person on the spring, The work results in stored elastic potential energy in the spring. If you look at the work energy theorem
$$W_{net} = \Delta K.E.$$ and since there is no KE change,
$$W_{net} = 0$$. But the net work is the sum of the work by the conservative forces (spring force) and non conservative forces (the 'we' pushing force). They must therfore be equal and opposite. We do so many joules of work, and the spring force does the same in the other direction.
Alternatively, you can use the total work energy method
$$KE_i + PE_{gravity}_i +PE_{spring}_i = KE_f + PE_{gravity}_f +PE_{spring}_f + W_{n.c.}$$
and since there is no kinetic or gravitational potential energy change, you get the same result for the work done by "we".