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Stretch of a Spring

  1. Oct 27, 2015 #1
    1. The problem statement, all variables and given/known data
    A block is sitting on a ramp with a spring behind it, which we are supposed to imagine we are holding. The mass of the block is 20 kg, the angle of the ramp is 30 degrees, and the spring constant is 500 N/m.

    1) You pull so the block is accelerating at 4 m/s^2 up the ramp. How far does the spring stretch?

    2) Now you are on a ramp with the following coefficients of friction: static friction coefficient=0.3 and the kinetic coefficient=0.2. You are initially holding the free end of the spring. You pull the spring so the block is accelerating at 4 m/s up the ramp, how far is the spring stretched?


    2. Relevant equations
    F=ma
    F=kx

    3. The attempt at a solution
    1)Can someone tell me if the way I solved for the stretch of the spring is correct? I'm not sure how to do it or how to incorporate acceleration into the equations in any other way.

    ΣFx=mgsinΘ=ma
    Fspring=mgsinΘ=kx
    mgsinΘ=kx
    ma=kx
    (20kg)(4m/s^2)=500 N/m)x
    x=0.16m

    2) ΣFx=mgsinΘ-fk=ma
    ΣFy=n-mgcosΘ
    n=mgcosΘ
    -I assume we use kinetic friction because the block is moving.
    fk=μ*n=0.2mgcos(30)=33.95N
    I'm not sure where to go from there to get x unless I set ΣFx=mgsinΘ-fk=kx but that doesn't seem right. But I did find in previous problems that Fspring=mgsinΘ=kx, so should I use ΣFx=kx-fk=ma ??? Or is this entirely wrong?
     
  2. jcsd
  3. Oct 27, 2015 #2

    billy_joule

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    Your first line is wrong, draw a free body diagram of the block.

    What you are saying with the equation below is that a single force acts in the x direction; the gravity force of the block.

    ΣFx=mgsinΘ=ma

    ie you've ignored the spring altogether. If we draw a FBD we'll see the sum of forces in the plane parallel to the incline is:

    ΣFx=Fx, gravity + Fx, spring = ma

    You've done the same thing for Q2)

    The spring extension will be greatest just before it moves, because μstatic > μkinetic.
    once the block is moving the extension is less. The question isn't very clear on which answer they want, ie should you consider the instantaneous extension before motion occurs? I would guess yes, that's why μs is given.
     
  4. Oct 27, 2015 #3
    https://flic.kr/p/A3aCsM for a picture like that, if that spring wasn't hooked to a wall and I was holding it and pulling it up the ramp like the question says, would the force of the spring be pointing back toward the spring since it's stretched? When the spring is stationary, there is not a spring force acting yes?
     
  5. Oct 27, 2015 #4
    So for 1) would i do ΣFx=mgsinΘ+Fspring=ma... turn fill in kx for Fspring to be ΣF=mgsinΘ+kx=ma and solve for x? In this case I get an x value of .134 m.

    I had another question of how far does the spring stretch while it is at rest and I put since it's at rest ΣF=0 and Fspring=mgsinΘ=kx, so Fspring=mgsinΘ= 98N, which I then used for Fspring=kx.. 98N=500N/n(x), which gave me x= 0.20 m for the stretch of the spring at rest.. So I don't think i'm doing the other question right if it's stretched less than it was at rest.. I guess I don't understand how to get the equation set up to solve for x
     
  6. Oct 27, 2015 #5

    billy_joule

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    I assume you are talking about the force the spring applies to the block? it'll be to the right, parallel to the ramp.


    If the block is stationary then there is no net force on the block so:
    ∑Fx = 0 =Fx, gravity + Fx, spring
    so
    Fx, gravity = - Fx, spring
    ie the spring applies a force to the block equal and opposite to gravity parallel to the incline.
    If the spring force were zero the block would be free to slide down the ramp.

    Show your working so we can find what went wrong.
     
  7. Oct 27, 2015 #6
    Ok, I'll back up to the previous question that wanted to know how far the spring stretched while it was at rest. I had this:
    Fspring=mgsinΘ= (20kg)(9.8m/s^2)(sin30)= 98 N
    Then I used Fspring=kx
    98 N= (500N/m) x
    x= 98/(500 N/m)= 0.20 m

    But, you're saying it should be mgsinΘ= -Fspring, which would give me a -98 N, which would give me an x= - 0.2 m. But that doesn't seem right to me? Am I supposed to use -Fspring=-kx, which would give me a positive value for x during this at rest point?

    Then for question 1 I had this: (which doesn't seem right)
    ΣFx=mgsinΘ+Fspring=ma
    (20kg*9.8m/s^2*sin(30))+Fspring= (20kg)(4m/s^2)
    I was putting in kx for the Fspring since Fspring=kx
    So,
    (20kg*9.8m/s^2*sin(30))+kx= (20kg)(4m/s^2)
    (20kg*9.8m/s^2*sin(30))+(500N/m)x= (20kg)(4m/s^2)
    98 N +(500N/m)x= 80
    which gives me an x of -.036....

    unless Fspring is supposed to equal -kx in this situation too, granted i even set it up right that far
     
  8. Oct 27, 2015 #7

    billy_joule

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    I found the following by summing the forces, I hadn't considered the direction (sign) of the forces.
    mgsinΘ= -Fspring

    If we apply a sign convention from the outset then this:

    Becomes this:
    ∑Fx = 0 = - Fx, gravity + Fx, spring
    so
    Fx, gravity = Fx, spring

    (where positive x is to the right, I was taught to draw a small arrow above the sigma to show what positive direction I had chosen for my sign convention)

    Your errors stem from not applying a consistent sign convention. Draw a labelled FBD showing the direction of all forces.
    Chose a positive direction and stick to it. Forces are vectors; they have magnitude and direction.
     
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