A block, a spring, and its distance travelled

[jnthn205]

Homework Statement

A(n) 2000g block is pushed by an external force against a spring (with a 15N/cm spring constant) until the spring is compressed by 15cm from its uncompressed length. The compressed spring and block rests at the bottom of an incline of 37degrees. Note: The spring lies along the surface of a frictionless ramp. Now, the external force is rapidly removed so that the compressed spring can push up the mass.
How far up the ramp will the block move before reversing direction and sliding back? The Block is not attached to the spring. Answer in units of cm.

[PLAIN]http://img405.imageshack.us/img405/43/springn.jpg

Homework Equations

Energy of the spring = 1/2kx^2
where Kinetic Energy = 1/2 kx^2 = 1/2mv^2

I think the distance = energy of the spring/force of gravity down the ramp.

The Attempt at a Solution

Using E=1/2kx^2 i found that the energy of the spring = 1687.5
and also, using Kinetic Energy = 1/2 kx^2 = 1/2mv^2. I found v = x*sqrt(k/m)
which would be 15sqrt(15/2000) = 1.3.
Is this correct so far?
I am kind of lost after this point

 

[Doc Al]

Don't forget about gravitational potential energy.

 

[jnthn205]

ok so i have the formula for distance = x^2 / [2(mg sin(theta)] L
which would be (15cm)^2 / [2(2000g)(980cm/s^2)(sin37)] 15 N/cm = .00143

This answer was wrong, but I think the formula is correct. I think I have my units messed up.
Can someone please input the correct units?

 

[Doc Al]

I recommend that you convert all quantities to meters and kilograms. You can always convert your answer back to the required units.

 

[jnthn205]

ok so
(.15m)^2 / [2(2kg)(9.8m/s^2)(sin37)] 1500 N/m = 1.43m.

But it wants the answer in cm, so 1.43 m = 143cm which was incorrect

 

[jnthn205]

i need help fast! This is due in 2 hours

 

[Doc Al]

ok so
(.15m)^2 / [2(2kg)(9.8m/s^2)(sin37)] 1500 N/m = 1.43m.

But it wants the answer in cm, so 1.43 m = 143cm which was incorrect

That's fine. Now express the answer as requested in the problem: How far up the ramp is that? (Where does the block start?)

 

[jnthn205]

143-45?? so 98cm?

 

[Doc Al]

143-45?? so 98cm?

No. You calculated how far up the ramp the block will go from its starting point--where is that starting point?

 

[jnthn205]

the starting point is when the spring is fully compressed right? So it starts at 45cm? Then it goes 143 more cm? So do I add them? 45+143?

 

[jnthn205]

or no, would you subtract the total spring length 45 subtracted byhow much it compressed 15? So 45-15 = 30. So the block starts at 30cm. So add 30cm to 143cm?

 

[Doc Al]

or no, would you subtract the total spring length 45 subtracted byhow much it compressed 15? So 45-15 = 30. So the block starts at 30cm. So add 30cm to 143cm?

If they want the block's position measured from the bottom, that would be right. But looking at your diagram, they define a distance "L" from the unstretched position of the spring. I assume that's what they want. What's L?

 

[jnthn205]

I don't know, I thought that's what I was solving for. How do I solve for L then?

 

[Doc Al]

I don't know, I thought that's what I was solving for. How do I solve for L then?

Use the diagram.

In your equation, you used L to represent the distance up the ramp the block moves. That's different than the L in the diagram. But you know where the block ends up--use the diagram to figure out the "L" that they want. (They want the position measured from the 45 cm mark from the bottom.)

 

[jnthn205]

so the block ends up 143+45cm down the ramp?

ugh I am so confused. Once you tell me I will know what your talking about

from what you are saying you want me to subtract 143 from 45, but I did that and you said it was wrong...so I am lost.

 

[Doc Al]

In post #11 you found the distance from the bottom. Since they want the distance from the end of the spring, just subtract 45 cm from that.

 

[jnthn205]

i found 143+30 so 183
so 183-45? so 138cm?

 

[Doc Al]

i found 143+30 so 183
so 183-45? so 138cm?

Right idea, but you made an arithmetic error.

 

[jnthn205]

173-45 =128? I only have 1 attempt left and I want to make sure this is correct before i input the answer

 

[Doc Al]

173-45 =128? I only have 1 attempt left and I want to make sure this is correct before i input the answer

That's what I would put.

 

[jnthn205]

Awesome thanks!
And to calculate the same thing, only this time with friction. I just plug in friction = .475 into the bottom of the equation from post 5?

so (1/2)(.15^2)/(2*9.8*sin(37))*1500*.475

that doesn't seem right, the distance would be greater than it is with no friction.
Or do I have to have a whole new equation for calculating in friction

 

[jnthn205]

Or is the new force when friction is involved = 1500N/m-(1500N/m*.475)=787.5 N/m

then use that in the equation? Like this:

(1/2)(.15^2)/(2*9.8*sin(37))*787.5

Is this correct?

 

[jnthn205]

Please doc don't give up on me now!

 

[jnthn205]

well thanks anyways...

 

[Doc Al]

Awesome thanks!
And to calculate the same thing, only this time with friction. I just plug in friction = .475 into the bottom of the equation from post 5?

so (1/2)(.15^2)/(2*9.8*sin(37))*1500*.475

that doesn't seem right, the distance would be greater than it is with no friction.
Or do I have to have a whole new equation for calculating in friction

With friction, you'll need to derive a new equation. Is .475 the coefficient of friction?

Write a new energy conservation equation.