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Horizontal spring

  1. Oct 22, 2010 #1

    afa

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    A 280000-kg wood block is firmly attached to a very light horizontal spring, as shown in the figure . The block can slide along a table where the coefficient of friction is 0.30. A force of 22 N compresses the spring .18m .
    If the spring is released from this position, how far beyond its equilibrium position will it stretch at its first maximum extension?

    I used 1/2 kx^2 = mew mg(x+D)
    D is the distance the block travels

    D= ((2*mew*mg)/k) -x

    didnt come up with the right answer and I dont know why..Please help??
     
  2. jcsd
  3. Oct 22, 2010 #2
    This is a conservation of energy problem.

    Initially you compress the spring, giving the system (block) potential energy of 1/2 kx^2 where x is the compression distance and k is the spring constant (which you can find). Then the spring is allowed to move, and that potential energy is converted to kinetic energy, and back to potential energy when the block fully stretches the spring out again. Without friction, E_initial = E_final, which means that x_initial = -x_final, the block moves the same distance but in the opposite direction from equilibrium, or the block as moved a total distance 2x. But now, we have friction. Friction is a disappative force that does work on the block as it moves. The energy lost due to friction is proportional to the total distance traveled (x_i + x_f). You should be able to solve the problem now.
     
  4. Oct 23, 2010 #3

    afa

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    i still come up with the same equation that doesnt work ???
     
  5. Oct 23, 2010 #4
    the change in total energy gives you the work from friction. No change in kinetic energy (the body is initially and finally at rest) so the change of its potential energy will be equal to the work of friction...
     
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