Work done on an object in a spring pendulum

[Faye716]

Homework Statement

The drawing shows three situations in which a block is attached to a spring. The position labeled "0 m" represents the unstrained position of the spring. The block is moved from an initial position x0 to a final position xf, the magnitude of the displacement being denoted by the symbol s. Suppose the spring has a spring constant of k = 45.0 N/m. Using the data provided in the drawing, determine the total work done by the restoring force of the spring for each situation. In the case of zero put your result as "+0".
Here is the link to the picture: http://www.webassign.net/cj8/10-p-025.gif

3. The Attempt at a Solution
w = fd
w = -k*x*d
w= 45 N/m * 2m *x
I don't know what to put in for x since the elongation of the spring changes... Do you put in the average elongation? (This is for part a of the picture)

 

[gneill]

Hi Faye716,

Welcome to Physics Forums!

Note that the force changes continuously as the spring changes length, so there's no one value for x that you can use. What do you do when you need to sum the contributions of a function of x between starting and ending limits of x?

 

[Faye716]

I think I'm supposed to do something with the integral but I don't know how to find it. I haven't taken calculus yet and we've just gone over it briefly in physics.

 

[gneill]

I think I'm supposed to do something with the integral but I don't know how to find it. I haven't taken calculus yet and we've just gone over it briefly in physics.

Writing and solving the integral would be the slick way to solve the problem. You want to integrate $F \cdot d$, where the "$\cdot$" represents the dot product since force direction versus change in distance matters.

The alternative is to use the formula for the potential energy stored in a spring and to analyze the system for the start and end position of the block, then apply the work-energy theorem.

 

[scottdave]

If you don't know Calculus, this may help. Think of a plot of a straight line, with the vertical axis Force, and the horizontal is distance. First, think of the work of a constant, unchanging force, which would plot out as a horizontal line. This makes a rectangle in a Force/distance plot. The area of a rectangle is F*x.

The spring plots as a straight line, but not horizontal. It starts at the origin, and has a slope of k (45 N/m). The area under this triangle is the work done. So work in this case is (1/2)(Force)(x). But force = k*x, so the Work formula for a spring is (1/2)*k*x^2. I hope this helps.