1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Springs: Speed when released & distance traveled.

  1. May 19, 2012 #1
    --------------------------------
    Question:
    --------------------------------
    A 1.6 kg block is pushed into a spring with spring constant of 1.0 x 10^3 N/m. The spring is compressed a distance of 2.0 cm [0.02m], and the block is released from rest in the horizontal direction.

    (a) Calculate the speed of the block as it passes through the equilibrium position at x = 0, if the surface is frictionless.

    (b) How far does the block travel before coming to rest? Assume a constant friction force of 4.0 N and the same initial conditions as before.
    --------------------------------
    Formula I used for (a)

    v = √[k/m (A^2 - x^2)]

    v = √[1000/1.6 (0.02^2 - 0^2)]
    v = 0.5m/s

    --------------------------------
    Not sure about (a), But I have no clue as to (b)..

    Should I used the work formula?
    Work (J) = Force x distance
     
  2. jcsd
  3. May 19, 2012 #2
    Part a is correct.

    Hint: Use Newton's second law followed by kinematic equation relating velocity, acceleration, and distance. Or use work energy relationship.
     
  4. May 19, 2012 #3
    This should work for (b):

    Conserving energy:
    (1/2)×k x2 = f * d

    where, k is spring constant, x is the compression of the spring, f is the frictional force, d is the distance traveled by the block before coming to a halt.
     
    Last edited: May 19, 2012
  5. May 20, 2012 #4
    This gave me the same result [5cm, 0.05m] as one of the ways I tried it originally:

    [v^2 = u^2 + 2ad]
    Where:
    a = F/m = 4N / 1.6kg = 2.5ms^-2
    v = 0.5ms^-1
    u = 0ms^-1

    (0.5)^2 = (0) + 2 (2.5) d
    0.25 = 5d
    d = 0.25/5
    d = 0.05m [5cm]

    -------------------------------------------------
    Your method:
    (1/2)×k x^2 = Ff * d
    Where:
    Ff = 4N
    k = 1000
    x = 0.02

    (1/2)× (1000) (0.02^2) = 4 * d
    500 x 0.0004 = 4 * d
    0.2 = 4 * d
    d = 0.2 / 4
    d = 0.05m [5cm]
     
  6. May 20, 2012 #5
    firstly, you put incorrect values into the formula v2=u2+2ad
    v is the final velocity, u is the initial velocity. the final velocity has to be 0, as the block stops due to frictional force. secondly, you have taken the velocity that you obtained from (a), which is the velocity of the block if frictional force was not acting. also, the acceleration should be negative, because the block is decelerating due to the frictional force. also,

    if you want to get the answer using your method, first find the velocity of the block where it leaves the spring, you can do this using energy conservation. now you can apply v2=u2+2ad
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Springs: Speed when released & distance traveled.
Loading...