Work & Energy (Spring and moving mass)

[CaptainSFS]

Homework Statement

http://pyrofool.googlepages.com/picspring.gif

A block of mass m = 3.5 kg rests on a frictionless floor. It is attached to a spring with a relaxed length L = 5 m. The spring has spring constant k = 12 N/m and is relaxed when hanging in the vertical position. The block is pulled d = 4 m to one side. In this problem, the block is always constrained to move on the floor (i.e. it never leaves the floor).

By what amount is the spring extended?
I found this to be 1.403 m.

What is the potential energy stored in the spring?
I found this to be 11.8125 J.

The block is released but is constrained to move horizontally on the frictionless floor. What is the maximum speed it attains?
I found this value to be 2.598 m/s.

Let's change the problem a bit. When the spring is vertical (hence, unstretched), the block is given an initial speed equal to 1.4 times the speed found in the last part.

How far from the initial point does the block go along the floor before stopping?
I found this to be 4.848 m.

What is the magnitude of the acceleration of the block at this point (when the spring is stretched farthest)?

This is the last part I cannot figure out.

Homework Equations

The Attempt at a Solution

I tried taking the KE (23.1526) that I found and tried to divide it by the distance (4.848m) (W=Fxd). And then I divided this force by the mass thinking it would give me the acceleration, and that did not work. I do not know what to do now. =/

Any help would be very much appreciated. thanks. :)

 

[Doc Al]

What forces act on the block?

 

[CaptainSFS]

the force of the push, and the force of the spring?

 

[Doc Al]

the force of the push, and the force of the spring?

At the point in question--when the spring is stretched farthest--there is no longer any "push" to worry about. (The "push" was just to start it moving.) The only horizontal force is due to the spring. Figure out what that is and use Newton's 2nd law.

 

[CaptainSFS]

Alright, So I set up the force to equal -k*x (-12*4.848) and then divided this answer by 3.5kg (F/m=a). That doesn't yield the correct answer either.

 

[Doc Al]

How much does the spring stretch? (Not 4.848 m--that's how far the block moves horizontally.) What's the horizontal component of the spring force?

 

[CaptainSFS]

yeah i tried that too. I found the distance the spring stretched to be 1.9644m. (25+4.848^2=(sqrt(answer))) I then multiplied that by -k, and divided by mass, and still not the correct answer.

 

[Doc Al]

Again: What's the horizontal component of the spring force?

 

[CaptainSFS]

I'm not sure how I find the horizontal component of the spring force.

sorry, I'm really trying to wrap my head around this, I'm just finding it difficult right now.

 

[Doc Al]

I'm not sure how I find the horizontal component of the spring force.

You find its horizontal component just like you'd find the horizontal component of any other vector. The spring force is directed along the spring. The triangle dimensions are known. Use a little trig.

 

[CaptainSFS]

alright, thx. I was trying to do that last night, but gave up when I thought all i had was one side. I just used a previous triangle to find the angle, and then used that to find x. thanks for your help, I really appreciate your input. :)

 

[r34racer01]

Hi I have a very similar question. How did you find the maximum speed it attains?

 

[Doc Al]

How did you find the maximum speed it attains?

Using conservation of energy.

 

[r34racer01]

Using conservation of energy.

I pretty lost can you explain that a little more?