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Homework Help: Mass colliding with a spring

  1. Oct 3, 2008 #1
    1. A 2 kg mass slides along a horizontal surface with coefficient of friction U(K) = 0.30. The mass has a speed of 1.3 m/s when it strikes the horizontal spring with constant k=120 N/m. (a) By what distance does it compress the spring(take friction into account). The spring will recoil and push the mass back, losing contact with the mass at its natural length. (b) What is the speed of the mass when it loses contact from the spring? (c) how far will it travel?

    2. Relevant Equations:

    F(spring) = -kx
    F(k) = U(k)mg
    Kinetic Energy = (1/2) mv^2
    Potential Energy = (1/2)kx^2

    3. So far, I have come up with the equation that

    F(net) = F(mass) - ((kx) + (U(K)mg))

    However, I could not figure out how to get the force of the mass since I was only given the velocity, and not the acceleration of the mass. I also thought that once the spring is at the most compressed spot, it will have all potential energy, and no kinetic energy. This is how I would solve question (b) if there were no friction. Since there is friction, this is not a conservative force, so I cannot use the K + U = constant equation. I have no idea where to go with this problem :-/
  2. jcsd
  3. Oct 3, 2008 #2


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    Whilst you are indeed correct that K + U =const. doesn't apply here, you can still use conservation of energy if treat the compression and extension parts of the collision separately.
  4. Oct 3, 2008 #3
    This does help! Thank you! However, I am stuck again. . .
    After seeing that the energy is conserved on the compression and the extension, I thought to make this equation:

    K = U + int( F(k) )

    I thought that taking the integral of the force of friction would give me the energy that is expended into friction. However, after plugging in the numbers, I came up with

    x= .22, -.13

    When I use these numbers for part (b) :

    U = (1/2)(120)(.22)(.22)
    = 2.904

    K- int( F(k) ) = U
    (1/2)(2)(v^2) = 2.904 + (2*9.8*.3*.22)
    v = 2.05 m/s

    This does not make any sense because energy is lost in this system to friction, and the velocity that it hits the spring with is smaller than 2.05. I do not know what I have done wrong. Actually, I am not sure if it is mathematically correct to take the integral of Friction to find the energy used in it, but that is how you find potential, so I thought that would work for friction as well.
  5. Oct 3, 2008 #4


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    Your over-complicating the problem by thinking about integration. Since the frictional force isn't a function of position, you can just use the elementary definition of work don by a force (namely the magnitude of the force multiplied by the distance traveled).
  6. Oct 11, 2008 #5
    just a bump to this thread, wanting to verify:

    You set the K.E. and the U(spring) equal to each other i.e.

    (1/2)mv^2 = (1/2)kx^2

    because of the conservation of energy? and this is why you don't need the force?

    EDIT: my problem is on a frictionless plane.
  7. Oct 12, 2008 #6


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    Yes, if there are no dissipative forces (e.g. friction) present nor any external forces, then the total energy of the system must remain constant.
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