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Homework Help: Spring Simple Harmonic Motion problem

  1. Jan 8, 2007 #1
    Hey, my teacher gave me these two problems for Homework and me nor anybody else in my class has no idea how to do them, any help would be greatly appreciated.

    1. a 15.0kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and readch a speed of 5m/s in .5s. In the process, the spring is stretched by .2m. The block is then pulled at a constant speed of 5m/s, during which time the spring is stretched by only .05m. Find (a) the spring constant of the spring and (b) the coefficient of kinetic friction between te block and the table.

    Now i used F applied=-kx with F applied=ma and x=displacement of the spring i got the spring constand to be equal to -750. That is the farthest i got, i dont know if its right or even how to begin part b.

    2.A 30.0kg blcok is resting on a flat horizontal table. ON top of this block is resting a a 15.0kg block, to which a horizontal spring is attached, as the drawing illustrates. The spring constant of the spring is 325N/m. The coefficient of kinetic friction between the lower block and the table is .6, while the coefficient of static friction between the two blocks is .900. A horizontal force is increading in such a way as to keep the blocks moving at a constant speed. At the point where the upper block begins to slip on the lower block determine (a) the amount by which the spring is compressed and (b) the magnitude of the force F.

    I dont even know where to start any help would be greatly appreciated.
    Last edited: Jan 8, 2007
  2. jcsd
  3. Jan 9, 2007 #2


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    Homework Helper

    You got the spring constant right, although it is not negative. Regarding part (b), which forces are acting on the block during its constant velocity? Further on, the question which is always asked, what does constant velocity imply?
  4. Jan 9, 2007 #3
    Constant velocity would imply that there is no acceleration and therefore the force applied and the friction would be equal. I do not however know how to get the coeficiant of friction. So would it be Force-Friction=0=kx?
  5. Jan 9, 2007 #4


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    Your equation does not make sense. It would make sense it you wrote something like ' Force - Friction = 0 '. The net force must equal zero, that part was good. Now, what does the force of friction equal?
  6. Jan 9, 2007 #5
    I'm not sure this is where i get confused. If Force-friction= 0 then wouldnt the force of friction have to be equal to the force. But since there is no acceleration how do you find the force. and kx canot be equal to 0
  7. Jan 9, 2007 #6


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    As radou said, constant acceleration (i.e. a = 0) implies that the net force (i.e. the vector sum of all the forces) must be zero. So, the first step is to identify all the forces acting on the block. Since, we are only concerned with forces in the horizontal plane (friction etc) we can ignore any forces acting in the vertical plane.
  8. Jan 9, 2007 #7
    well since the acceleration is zero that means that the friction would be equal to the force acting positively on the block. But how do i find what the friciton is equal to? it would be mu Force normal, but what is that equal to.
  9. Jan 9, 2007 #8

    Doc Al

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    In solving this problem you'll need to set up two equations, one for each scenario: constant acceleration & zero acceleration.

    This is incorrect. In Newton's 2nd law you need to use the net force, not just the force applied by the spring. You forgot to include friction.

    "mu" is one of the two unknowns you are solving for, but what is the normal force? (What forces act vertically on the block?)

    Again, to solve this problem and find "k" and "mu", apply Newton's 2nd law twice: once for the constant acceleration case, once for the constant velocity case. Solve those two equations together and you'll get your answers.
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