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RJLiberator
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Homework Statement
The pulley in the figure has radius R=0.160m and a moment of Inertia I_p = 0.560 kg*m^2. The rope does not slip on the pulley rim. As m_4 (4kg block) drops due to gravity, m_2 (2kg block) is dragged up a ramp inclined at theta = 65 degrees. The coefficient of kinetic friction for the block/ramp is 0.100. Use energy methods to calculate the speed of m_2 after m_4 has dropped 5.00 meters. (Hint the normal force on m_2 is m_2gcos(theta) and change in height is not the same for m_2 and m_4 although they are related to each other).
Homework Equations
The Attempt at a Solution
So first, decrease in PE is equal to increase in KE.
4*5*9.8 = 196 J.
The frictional work done is 1.96*cos65*5 = 9.8cos(65)
The force parrallel is 19.6sin(65)*5 = 98sin(65)
Final KE of the system = 196-9.8cos(65)-98sin(65)
Final KE's of the syem
M_4 = 1/2*4*v^2
M_2 = 1/2*2*v^2
Pulley = 1/2*I*w^2 ==> using w=v/R ==> 10.938v^2
Total KE = 13.938v^2
So:
13.938v^2 = 196-9.8cos(65)-98sin(65)
Which results in a 2.72 m/s
Should I have used the coefficient of friction instead of doing what I did? :O
Is there anything wrong with this reasoning?