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Sliding Block

  1. Mar 31, 2014 #1
    A 1.0 kg block is held in place against a spring with spring constant k = 100 N/m (see the figure, below) by a horizontal external force. The spring is compressed 20 cm. The external force is removed, and the block is projected with some (horizontal) velocity upon separation from the spring. The block descends a ramp with height h = 2.0 m. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction over this section is 0.30. The block slides through this rough section a distance S, and comes to a stop at point D.
    What is the sliding distance S?

    Lokk6AE.png

    Can someone please tell me if my below attempt is the correct answer?

    Find velocity after reaches bottom of hill (by conservation of mechanical energy)
    ( 1/2 * k * x2 ) + ( 1/2 * m * g ) = 1/2 * m * v2
    (1/2 * 100 * .20) + (1/2 * 1 * 9.8) = 1/2 * 1 * v2
    14.9 = 1/2 * v2
    v ≈ 5.4589 m/s

    Find the acceleration once it hits the rough spot
    μmg = Fnet
    0.30 * 1 * 9.8 = 2.94 N

    2.94 N = m * a
    2.94 N = 1 * a
    a = -2.94 m/s2

    Kinematics to find distance of S
    vf2 = vi2 + 2ax
    0 = 5.45892 + ( 2 * -2.94 * x)

    x = 5.068 m
     
  2. jcsd
  3. Mar 31, 2014 #2

    Simon Bridge

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    KE bottom of hill -
    check - gravitational potential energy, and shouldn't the 0.20m be squared?

    Best practice:
    - finish the algebra before subbing in the numbers.
    i.e. ##-\mu m g = ma\implies a=-\mu g## ... then substitute.

    - use the variable names from the problem in your equations.
    i.e. the distance slid is S not x.
     
  4. Mar 31, 2014 #3
    Awesome, thank you for the tips and catching my mistake.
    Could you review my corrections and let me know if my answer now seems to be correct?

    Find velocity after reaches bottom of hill (by conservation of mechanical energy)
    ( 1/2 * k * x2 ) + ( 1/2 * m * g ) = 1/2 * m * v2
    (1/2 * 100 * .202) + (1/2 * 1 * 9.8) = 1/2 * 1 * v2
    6.9 = 1/2 * v2
    v ≈ 3.7148 m/s

    Find the acceleration once it hits the rough spot
    -μmg = ma
    a = -μg
    a = 0.30 * 9.8 = -2.94 m/s2

    Kinematics to find distance of S
    vf2 = vi2 + 2ax
    0 = 3.71482 + ( 2 * -2.94 * S)

    S ≈ 2.3469 m
     
  5. Mar 31, 2014 #4

    Simon Bridge

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    ... check this:
    "1/2 * m * g" is "half the weight" - a force. All the other terms are energy.

    You should get used to scanning your equations to see that the dimensions make sense.
    Please note: I won't check your arithmetic.
     
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