Help: Kinetic Energy and Friction  A block pulled by another on a pulleyby aarno Tags: conservation of ener, friction, kinetics 

#1
Jul1810, 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 



#3
Jul1810, 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 deltax equals 2.47 as provided in the problem. Therefore, W = F * deltax = (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. 


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