Speed when Spring is Stretched

[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?

 

[Doc Al]

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

(But considering energy is the right idea.)

 

[Manni]

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

(But considering energy is the right idea.)

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.

 

[Doc Al]

I use it to find the amplitude

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

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

 

[maCrobo]

Yes, you can rearrange for v. :D

 

[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

 

[Doc Al]

Another hint. You'll use this:

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

But what's E? What's putting energy into this system?

 

[maCrobo]

what's your wrong answer? and what's the right one?

 

[Manni]

what's your wrong answer? and what's the right one?

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

 

[Doc Al]

Yes, you can rearrange for v. :D

No you can't.

 

[maCrobo]

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

 

[Doc Al]

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

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

 

[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?

 

[JHamm]

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

 

[Manni]

Oh, I got it!

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

 

[Doc Al]

Oh, I got it!

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

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

 

[drawar]

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