Compressed spring with work

[endeavor]

The force required to compress an imperfect horizontal spring an amount x is given by F = 150x + 12x³, where x is in meters and F in Newtons. If the spring is compressed 2.0m, what speed will it give to a 3.0 kg ball held against it and then released?

I know how to integrate F(x) to get the work done, and with that I could use the work-energy theorem to find the speed. But do I take the lower limit of the integral as x = 2.0 and the upper limit as x = 0?
But then the integral would be negative and 1/2mv² can't be negative...

 

[Mindscrape]

Are you finding the work function by solving $$F = -\frac{dU}{dx}$$? I would integrate from x=0 to x=x to find the general solution, but if you wanted to put in 2 right away then that would work. The negative thing depends on where you define your zero potential.

 

[m2003]

Yes, you are correct in your approach. To find the work done by the spring, you would need to integrate the force function from the compressed position (x = 2.0m) to the uncompressed position (x = 0). This would give you a negative value for work, as the force is acting in the opposite direction of displacement. However, when using the work-energy theorem, you would use the magnitude of this negative work value, as the work done by the spring would result in an increase in kinetic energy for the ball.

So, using the work-energy theorem, we can write:

W = 1/2mv^2

Where W is the work done by the spring and v is the final velocity of the ball. Rearranging this equation, we get:

v = √(2W/m)

Substituting the value of work (negative magnitude) and mass (3.0 kg), we get:

v = √(-2(∫F(x)dx)/m)

Integrating the force function, we get:

v = √(-2(75x^2 + 4x^4)/3)

Substituting the values of x (2.0m) and solving, we get the final velocity of the ball as:

v = 5.7 m/s

So, when the compressed spring is released, it will give a speed of 5.7 m/s to the 3.0 kg ball. This answer may seem counterintuitive as we are using a negative value for work, but it is important to remember that work done by a force in the opposite direction of displacement will result in a decrease in kinetic energy, hence the negative sign.