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Draw a FBD, calculate tension in the string, determine mass

  1. Aug 4, 2017 #1
    1. The problem statement, all variables and given/known data

    A box slides on a smooth (frictionless) horizontal surface, 1.50 m above the floor, at a speed of 2.00 m/s. The box then slides down a ramp that makes an angle of 36 deg with the horizontal and has a coefficient of kinetic friction equal to 0.430. After reaching the end of the ramp, the box continues to slide horizontally on a smooth ground. Use the law of conservation of energy to calculate the final speed (v) of the box as it slides on the floor.

    2. Relevant equations

    KEi + PEi = KEf + PEf

    and

    KEi + PEi + Wnc = KEf + PEf

    3. The attempt at a solution

    I broke it into three segments. I first did the flat floor above the ramp. Since it is all conservative forces acting, I used KEi + PEi = KEf + PEf. Furthermore, down the ramp i defined as negative and south is negative.

    Therefore,

    1/2m*vi^2 + mghi = 1/2m*v^2 = mgh

    In the above equation, the height initial (hi) and the final height (h) are both the same. So the mgh will cancel leaving

    1/2m*vi^2 = 1/2m*v^2

    After simplifying,

    vi^2=v^2 --> Since vi is -2.00 m/s (it is going down in the direction towards the ramp, which is to the left_

    (2.00)^2 = v^2
    4=v^2
    v=-2.00 m/s --> This will now become the initial velocity for the segment involving the ramp.

    Since the ramp has friction, we need to consider work by the non-conservative forces -->

    KEi + PEi + Wnc = KEf + PEf

    1/2m*vi^2 + mghi + Wnc = 1/2m*v^2 + mgh

    Where: The initial velocity (vi) = -2.00 m/s, the initial height (hi) is 1.5m, the final height (h) is 0, and wnc= dfcos(theta).

    1/2m*(-2)^2 + m(9.8)(1.5) + d*friction*cos(theta) = 1/2m*v^2 + m(9.8)(0)

    Therefore, 16.7m - d*uk*N = 1/2m*v^2

    The distance (d) is: sin(36) = 1.5 --> d = 2.5519m

    16.7m - (2.5519m)(0.430)*N = 1/2m*v^2

    16.7m - 1.0973N = 1/2m*v^2

    I use a force diagram of the box on the ramp, and I see that N=w perpendicular --> N= mgcos(36)

    33.4m - mgcos(36) * 2.1946 = v^2

    33.4 - 17.3995 = v^2

    16=v^2

    v= +/- 4.00 m/s --> I use negative value, again because I defined down the ramp as negative direction for x

    Finally, just to show that this is the final velocity of the box, I take the segment of the box just hitting the ground where only conservative forces act, to show that the final velocity as it is sliding on the floor is -4.00 m/. This question asks one for speed, which is scalar, so the answer is 4.00 m/s. Is this the correct logic for a problem like this? Could someone verify the answer?

    Thank you in advance.
     

    Attached Files:

  2. jcsd
  3. Aug 4, 2017 #2

    scottdave

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    The logic looks fine for the ramp problem. There were a couple of typos, like when you had: 33.4m - mgcos(36) * 2.1946 = v^2

    It should be 33.4m - mgcos(36) * 2.1946 = mv^2.
    Also, your picture is for a totally different problem. So which problem are you asking about?
     
    Last edited: Aug 4, 2017
  4. Aug 4, 2017 #3
    Thank you, i got 4 but another friend of mine got 7? Do you know if my math is right?

    For the picture I attached I got the answers to be T= 0.93 N

    I got the mass of cylinder to be 3.72 kg..

    Does that seem right? Furthermore, for the FBD for the block it should only be Tension upwards and weight downwards? And then the cylinder should have both tension and weight downwards? Does that make sense?

    Thank you again!
     
  5. Aug 4, 2017 #4

    scottdave

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    So the cylinder turns and has a moment of inertia. Do you know how to handle those type of problems?
     
  6. Aug 4, 2017 #5
    Yup. I got the mass from that moment of inertia using torque = I * the angular acceleration.

    Do you know how the free body diagram should look for the cylinder? Is it tension and weight pointing down?
     
  7. Aug 4, 2017 #6

    scottdave

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    OK. Yes tension is down, weight is down, and reaction force at the axle is up.
     
  8. Aug 4, 2017 #7
    Thank you so much. Did you happen to get the same tension and mass by any chance? Some reassurance would be awesome. LOL
     
  9. Aug 4, 2017 #8

    scottdave

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    I also came up with 4 m/s.
    I arrived at the same values for the tension and the cylinder mass. I wasn't sure if you need to know the radius, but it turns out that the radii cancel, so two cylinders of vastly different size but the same mass (say one made of styrofoam and the other made of lead) would allow the falling mass to behave the same.
     
  10. Aug 5, 2017 #9
    Thank you so much! I thought the same thing as you, which is very interesting!
     
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