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Find the work done on the block by the spring

  1. Mar 18, 2004 #1
    A 5kg block is held against a spring of force constant 20N/cm, compressing it 3cm. The block is released and the spring extends, pushing the block along a horizontal surface. The coefficient of friction between the surface and the block is 0.2.

    a) Find the work done on the block by the spring as it extends from its compression to its equilibrium position.
    b) Find the energy dissipated by friction while the block moves the 3cm to the equilibrium position of the string.
    c) What is the speed of the block when the spring is at its equilibrium position?
    d) If the block is not attached to the spring, now far will it slide along the surface before coming to rest?
     
  2. jcsd
  3. Mar 18, 2004 #2

    HallsofIvy

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    If you are not able to do anything with these problems then you have much worse problems than we can do anything about! At least, show us what you have tried so we will know how to help you.

    a) Find the work done on the block by the spring as it extends from its compression to its equilibrium position.

    Do you know any formula for work done by a spring?

    b) Find the energy dissipated by friction while the block moves the 3cm to the equilibrium position of the string.
    This is easier than a) because the force is a constant. How do you find the work done by a constant force moving an object 3 cm? Do you know how to find the friction force given the "coefficient of friction"?

    c) What is the speed of the block when the spring is at its equilibrium position?
    One way to do this (the hard way!) is to calculate the total force (spring force and friction force) at each instant, use that to calculate the speed and position as functions of time, solve for the time that the block is at the equilibrium position and use that to find the speed of the block.

    Much simpler is to find the potential energy in the compressed spring (if this isn't a calculus course, you have a formula for that). You calculated the work done by friction on the block in (b). Subtract that from the potential energy you just found. The rest is converted into kinetic energy:(1/2)mv2.
     
  4. Mar 18, 2004 #3
    As you may see that I am still working on the previous ones and I did post my work. I am really lost in the concept of the work systems.
     
  5. Mar 19, 2004 #4
    should I take friction into concideration on b?
    can I use mgh=1/2k(x^2)? I am really lost with this one.
     
    Last edited: Mar 19, 2004
  6. Mar 19, 2004 #5

    HallsofIvy

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    "should I take friction into concideration on b? "

    Now that's a very strange question! (b) says "Find the energy dissipated by friction". Yes, you have to take friction into consideration in order to find the energy dissipated by friction!

    "Work" is "force times distance" for a constant force and, for a variable force, f(x), becomes integral of f(x) dx. In the case of a spring with spring constant 20 N/cm, f(x)= -20 x.
    [tex]\int_0^3 -20xdx= (-10)x^2\|_0^3= -90 Joules[/tex].
    (That's your formula "mgh=1/2k(x^2)" although "work" is not always "mgh"- that's the work done in lifting something a height h and there is no height in this problem). That's the work done in compressing the spring 3 cm and so the work done on the block by the spring as it "uncompresses" is 90 Joules.

    The block has mass 50kg and so weight 50g N. The coefficient of friction is 0.2 so the friction force is (0.2)(50g)= 25g N. The work done moving the block .03 m is (0.03)(25g)= 0.75g or approximately 7.35 J.

    Since 7.35 J goes into pushing against friction, the rest of the work, 90- 7.35= 82.65 J changes into kinetic energy. The kinetic energy of the block is (1/2)mv2= 45v2= 82.65 so v2= 1.84 and v= 1.36 m/s at the equilibrium point. ("20 N/cm" is one heck of a strong spring!)

    If the block is not attatched to the spring, then once it has passed the equilibrium point, the spring no longer acts on it. Past that point the only force is the 25g= 245.25 N friction force. The total energy of the block at the equilibrium point is 82.65 J and the work done by the friction stopping the block must equal that. Letting x be the distance past the equilibrium point the block slides, 245.25x= 82.65 so x= 82.65/245.25= 0.35 m or 34 cm.
     
    Last edited: Mar 19, 2004
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