# Speed when Spring is Stretched

1. Jan 7, 2012

### Manni

A 1.50 kg block is at rest on a table and is attached to a horizontal spring with a spring constant of 19.6 N/m. The spring is initially not stretched. A constant force of 20 N horizontal force is applied to the object causing the spring to stretch. Determine the speed of the block when it has stretched 0.300 m. Assume the table is frictionless.

The way I did it:

E = (mv^2)/2 + (kx^2)/2
1/2kA^2 = (mv^2)/2 + (kx^2)/2

Do I simply rearrange for speed?

2. Jan 7, 2012

### Staff: Mentor

What about the 20 N force? Where does that come in?

(But considering energy is the right idea.)

3. Jan 7, 2012

### Manni

I use it to find the amplitude

F = -kx
20 = -kA
-A = 20/k

Negative sign can be omitted since amplitude is treated in a A/-A fashion.

4. Jan 7, 2012

### Staff: Mentor

No, that's not quite right. The force is continuously applied. Instead, consider the work done by that force.

5. Jan 7, 2012

### maCrobo

Yes, you can rearrange for v. :D

6. Jan 7, 2012

### Manni

Great, but when I did that I still got the wrong answer.

The answers I have to choose from are:

4.47 m/s, 70.6 m/s, 44.7 m/s, 2.61 m/s, and 8.96 m/s

Last edited: Jan 7, 2012
7. Jan 7, 2012

### Staff: Mentor

Another hint. You'll use this:
But what's E? What's putting energy into this system?

8. Jan 7, 2012

9. Jan 7, 2012

### Manni

I got 3.52 m/s, and it's not even offered as one of the choices.

10. Jan 7, 2012

### Staff: Mentor

No you can't.

11. Jan 7, 2012

### maCrobo

Then, I'm getting curious. E = (mv^2)/2 + (kx^2)/2 can't he simply rearrange for v? O_O why?

12. Jan 7, 2012

### Staff: Mentor

Before solving for v, you must first find the correct expression for the energy, E.

13. Jan 7, 2012

### maCrobo

Think about this: the spring and the mass can be thought as a system, then you apply a force F on it... so what happens to the energy?

14. Jan 8, 2012

### JHamm

Work energy theorem seems like the simplest way to solve it.

15. Jan 11, 2012

### Manni

Oh, I got it!

You need to sub in (A-0.300) meters for x, and not 0.300 m!

16. Jan 12, 2012

### Staff: Mentor

At the point in question, the displacement from equilibrium x is given as 0.300 m. (Forget about the amplitude; this is not SHM.)

17. Jan 12, 2012

### drawar

Elastic P.E + K.E = Total Energy = Work done when the mass travels 0.3m under a force whose magnitude is 20N