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Help: Kinetic Energy and Friction - A block pulled by another on a pulley

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aarno
#1
Jul18-10, 04:28 PM
P: 2
In the system the following diagram shows, the block M (mass of 15.65 kg) is initially moving to the left with a speed of 2.26 m/s. The mass of m is 8.26 kg and the coefficients of friction are μs = 0.411 and μk = 0.304. The string is massless and the pulley is massless and frictionless. How fast (m/s) will M be traveling when m has fallen through a height h=2.47 meters?

diagram: http://ce.byu.edu/courses/univ/69482..._blockmass.jpg
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Doc Al
#2
Jul18-10, 05:10 PM
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Well, what do you think? What have you done so far?
aarno
#3
Jul18-10, 06:41 PM
P: 2
^

The pulley makes it so that the second mass exerts a force purely in the x direction, not affecting the normal force. I used left (or down, for the second mass), as the positive direction.
The force exerted by mass m is equal to its mass times gravity, and delta-x equals 2.47 as provided in the problem. Therefore,
W = F * delta-x = (8.26)(9.8)(2.47) = 199.942

The normal force is equal to the weight of mass M, so the force of kinetic friction:
fk = m*g*μk = (15.65)(9.8)(0.304) = 46.6245

Using the following equation, then plugging in the values:
0.5*m*vf^2 = 0.5*m*vf^2 - fk*d + W
0.5*15.65*vf^2 = 0.5*15.65*2.26^2 - 46.6245*2.47 + 199.942

This give vf = 3.993 m/s, while the correct response is 3.49 m/s.

Doc Al
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Jul18-10, 06:54 PM
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Help: Kinetic Energy and Friction - A block pulled by another on a pulley

Quote Quote by aarno View Post
The force exerted by mass m is equal to its mass times gravity,
That's incorrect. Note that the masses are accelerating. You'll have to solve for the tension in the rope.


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