- #1

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?