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Work Problem (Block with spring)

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data

    A spring (k=600 N/m) is at the bottom of an inclined plane that makes and angle θ=30° with the horizontal. The spring is compressed by 0.10 m and a block of mass M=2.0 kg is placed against it, as shown below. The block is then released from rest.

    a) Determine the speed of the block just as it loses contact with the spring.

    b) First assume that the plane is frictionless. How high up the incline ( L in the diagram) will the block travel before it stops momentarily?

    c) Next you repeat the experiment, but first you roughen up the block with sandpaper so that there is now friction between the block and the plane beyond where the block leaves the spring. If the coefficient of friction is now μ=0.2, how high up the incline (L in the diagram) will the block travel before it stops momentarily?


    For a picture of the diagram http://egp.rutgers.edu/115/115072nd1.pdf [Broken]

    it is the second problem...


    a) I said the Initial Uspring = Final KEf

    1/2k[tex]\Delta[/tex]x2=1/2mv2

    1/2(600n/m)(.10m)^2=1/2(2kg)v^2

    V = 1.73m/s ?

    For parts b and c, i am confused because they do not give you the L, distance the block slides, or the H (the height of the inclined plane)
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 16, 2009 #2

    rl.bhat

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    Since the angle of inclination is given, L and H are not required.
    Find out which force is preventing its motion in the upward direction. In other wards, what is the component of g along the inclined plane?
     
  4. Nov 16, 2009 #3
    Well, weight is mg and that equals (2kg)(-9.8)= -19.6N

    Fn is mgcos[tex]\theta[/tex] (2kg)(9.8)cos(30) which equals 16.9N
     
  5. Nov 16, 2009 #4

    rl.bhat

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    What happens to the other component weight which acts along the surface of the inclined plane?
    Then use the energy equation
    1/2*m*v^2 = F*s. where s is the displacement along the inclined plane.
     
  6. Nov 16, 2009 #5
    um, gravity's x component is

    sin30*(2)(-9.8)= -9.8N or would this be positive? Is it pointing the same direction the block is moving or away?

    1/2*m*v^2 = F*s

    1/2(2)(1.73)^2 = (-9.8N)s

    s= -0.31m

    okay, i'm thinkin the -9.8N is positive because you can't travel -0.31 m but 0.31m
     
  7. Nov 16, 2009 #6

    rl.bhat

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    Here f ad s are in the opposite direction. So F.S is negative. Hence s = 0.31 m.
     
  8. Nov 16, 2009 #7
    In my calculation of s, should I just disregard the negative sign then?

    for part c

    1/2mv^2 = mgh + Ff*s

    1/2mv^2 = mg(sin[tex]\theta[/tex]*s) + [tex]\mu[/tex]mgcos[tex]\theta[/tex]*s

    1/2(2)(1.73)^2 = (2)(9.8)(sin30*.31)+(0.2)(2)(9.8)cos(30)*s

    s=0.013m?
     
  9. Nov 16, 2009 #8

    rl.bhat

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    1/2(2)(1.73)^2 = (2)(9.8)(sin30*.31)+(0.2)(2)(9.8)cos(30)*s
    It is not correct.
    It should be
    1/2(2)(1.73)^2 = [(2)(9.8)(sin30*)+(0.2)(2)(9.8)cos(30)]*s
     
  10. Nov 17, 2009 #9
    Oh, you distribute out the s. Thanks. For part b though, can you explain to me how it is

    1/2*m*v^2 = F*s ?

    Would it not have mgh in it's final state? it did move to a higher place than it initially started right?
     
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