How Fast Does a Block Move on a Spring?

[Ronnin]

A 500g block is attached to a 60cm long massless spring with spring constant of 40N/m and is at rest with the back of the block at point A on a frictionless table. A horizontal force of 20N is applied to pull the block towards the right. What is the block's speed when the back of the block reaches point B which is 25cm to the right of point A?

Spring Potential=.5(40N/m(.0625m^2)=1.25J
Kinetic Energy=Fx=20N(.25m)=5J
ME=KE+PE = 6.25J=(.5kg(v^2))/2 + 1.25J
v=4.47m/s

Does this look right?

 

[Hootenanny]

A 500g block is attached to a 60cm long massless spring with spring constant of 40N/m and is at rest with the back of the block at point A on a frictionless table. A horizontal force of 20N is applied to pull the block towards the right. What is the block's speed when the back of the block reaches point B which is 25cm to the right of point A?

Spring Potential=.5(40N/m(.0625m^2)=1.25J

Where did this value come from?

 

[Ronnin]

Where did this value come from?

.25m^2, the distance from A to B

 

[Hootenanny]

.25m^2, the distance from A to B

Indeed, but this would not give you the initial potential energy of the system before the spring was released.

 

[Dorothy Weglend]

There are two points of interest in this problem. The point before the release of the spring (after the force is applied) and the point .25 m to the right of A. The energy at both of these points is equal. One will have only potential energy, the other will have a combination of potential and kinetic energy.

I think you meant to write 0.625 for your spring potential calculation, not 0.0625. As I read the problem 0.60 m is the equilibrium point of the spring, not the stretched length.

Dorothy

 

[Dorothy Weglend]

I wrote my reply before I saw Hootenanny's comment. So disregard my guess about the origins of the 0.0625 m. You do need to find the potential energy of the system before the spring is released.

 

[Ronnin]

The block has yet to be released. I'm assuming that it is asking for the speed of the block as it is being pulled along with the 20N force. I used the sum of work to pull the block along + the springs potential to give me the total ME for the system. Am I looking at this the wrong way?

 

[Hootenanny]

The block has yet to be released. I'm assuming that it is asking for the speed of the block as it is being pulled along with the 20N force. I used the sum of work to pull the block along + the springs potential to give me the total ME for the system. Am I looking at this the wrong way?

Ahh, my apologies, I misinterpreted the question.

 

[Ronnin]

For the second part of the question it asks.
When the block reaches B, you let it go back. How close does it get to the wall where the spring is attached.

Ki + Ui = Kf + Uf
6.25J=0+1/2(kx^2)
.3125m^2=x^2
x=.56m
it gets .29m close to the wall (Equilibrium+Stretch=.85-.56)

 

[Dorothy Weglend]

Yes. Me too. Sorry about that.

 

[Ronnin]

I appreciate and need all the help I can get. Hoot, Dorothy, thanks for taking the time to reply. I interpreted it exactly the same way you guys did the first time I looked at it. This was a test question I'm trying to correct. Does everything else look ok?

 

[OlderDan]

I appreciate and need all the help I can get. Hoot, Dorothy, thanks for taking the time to reply. I interpreted it exactly the same way you guys did the first time I looked at it. This was a test question I'm trying to correct. Does everything else look ok?

If I am interpreting the problem correctly, your first part is not correct. I take it to mean that the block is connected to the spring and initially at rest with the spring initially unstretched, so the energy of the block/spring system is initially zero. The 20N force is then applied to the block and you are looking for the speed after it hs moved 25 centemeters. The 20N is an additional force to the force applied by the spring. So looking at your original post, I see the following difficulty

A 500g block is attached to a 60cm long massless spring with spring constant of 40N/m and is at rest with the back of the block at point A on a frictionless table. A horizontal force of 20N is applied to pull the block towards the right. What is the block's speed when the back of the block reaches point B which is 25cm to the right of point A?

Spring Potential=.5(40N/m(.0625m^2)=1.25J
This is the energy stored in the spring when the block gets to 25cm. This looks good

Kinetic Energy=Fx=20N(.25m)=5J
20N is not the only force acting on the block. The spring is being stretched (or compressed) Not all of this work is going into kinetic energy of the mass. It is going into kinetic energy and the energy of the spring.

ME=KE+PE = 6.25J=(.5kg(v^2))/2 + 1.25J
ME is the increase (from zero) in energy of the mass/block system as a result of the 5J of work done by the 20N force.

v=4.47m/s
Not quite so fast.

Does this look right?

For the second part of the question it asks.
When the block reaches B, you let it go back. How close does it get to the wall where the spring is attached.

Ki + Ui = Kf + Uf
6.25J=0+1/2(kx^2)
.3125m^2=x^2
x=.56m
it gets .29m close to the wall (Equilibrium+Stretch=.85-.56)

If my interpretation in the first part is correct, you will need to adjust the total energy on the LHS and find a new x. But .85m is not the equilibrium length of the spring. That is the length of the stretched spring when you stop applying the 20N force.

 

[Dorothy Weglend]

For the second part of the question it asks.
When the block reaches B, you let it go back. How close does it get to the wall where the spring is attached.

Ki + Ui = Kf + Uf
6.25J=0+1/2(kx^2)
.3125m^2=x^2
x=.56m
it gets .29m close to the wall (Equilibrium+Stretch=.85-.56)

Thanks for your graciousness, Ronnin. Well, I'm not batting 1000 at this problem, but it seems to me that conservation of energy would imply that kx^2/2 would be the same at both positions of the spring, assuming that the extra 20N force is not present when you let the spring go back, of course.

Dorothy

 

[Dorothy Weglend]

If my interpretation in the first part is correct, you will need to adjust the total energy on the LHS and find a new x. But .85m is not the equilibrium length of the spring. That is the length of the stretched spring when you stop applying the 20N force.

Ummm. So I guess I was wrong again. Whoops. The x on the left wouldn't equal the x on the right, because of inertia, I guess? Once the force is removed, the spring continues to move a bit to the right, so you need to account for this extra kinetic energy in calculating the amount the spring is compressed, in other words?

Dorothy

 

[OlderDan]

Ummm. So I guess I was wrong again. Whoops. The x on the left wouldn't equal the x on the right, because of inertia, I guess? Once the force is removed, the spring continues to move a bit to the right, so you need to account for this extra kinetic energy in calculating the amount the spring is compressed, in other words?

Dorothy

The maximum distance left and right is equal, but as you say the mass is not finished moving right when the 20N force stops at 25cm. It contiunes to move right until all the energy is spring energy, then reverses and goes the same distance past equilibrium on the other side.

 

[Ronnin]

Wow, i didn't even think about the inertia. I'm going to confirm the exact parameters of the question because this seems a bit out of scope. Dan, if I am pouring 5J into this system then that is my total energy. Part goes into KE and part into PE? So only 3.75J is going into KE because the spring is absorbing the rest? Am I following this correctly?

 

[OlderDan]

Wow, i didn't even think about the inertia. I'm going to confirm the exact parameters of the question because this seems a bit out of scope. Dan, if I am pouring 5J into this system then that is my total energy. Part goes into KE and part into PE? So only 3.75J is going into KE because the spring is absorbing the rest? Am I following this correctly?

Yes. And that is of course going to keep things moving to the right for some distance after the force is gone.

 

[Ronnin]

V=3.87m/s. How does that look

 

[OlderDan]

V=3.87m/s. How does that look

Looks OK. Did you find the distance from the wall?

 

[Ronnin]

5J=1/2(.5kg)(3.87m/s)^2+1/2(40N/m)x^2
x=.2506m extra travel past the B point

But this doesn't seem right because U at this new distance of .5006m of stretch give me more than 5J of energy. Did I set this up wrong?

 

[OlderDan]

5J=1/2(.5kg)(3.87m/s)^2+1/2(40N/m)x^2
x=.2506m extra travel past the B point

But this doesn't seem right because U at this new distance of .5006m of stretch give me more than 5J of energy. Did I set this up wrong?

I'm not sure what you are doing in this calculation, but it looks to me like an accident of the numbers is leading you in the wrong direction. At the point where you have this kinetic energy, x = 25cm. You already know that. That is how you found the kinetic energy and the velocity at this point. Increasing the displacement until the mass stops increases the potential energy by 3.75J to a total of 5J.

Knowing that the total energy is 5J, and that at maximum displacement the velocity is zero gives you |x|_max = 50cm. The mass moves 50cm to both sides of equilibrium. It just happens that the mass moves an additional 25cm, doubling the distance the force was applied. If the force had stopped at 20cm, the mass would not just go another 20cm; it would still go almost 25 cm more.

 

[Ronnin]

I found my error and got the same answer you did. Thank you so much for your help on this.