A 4kg block resting on a horizontal ledge is attached to his right to a string that passes over a pulley and is attached to a hanging 2kg block. The coefficient of friction between the ledge and the 4kg block is 0.25. The pulley is a uniform disk of radius 8cm and mass 0.6kg. a) Find the speed of the 2kg block after it falls from the rest a distance of 2.5m. b) What is the angular velocity of the pulley at this time?(adsbygoogle = window.adsbygoogle || []).push({});

Delta K + Delta U = 0

Because Ki=Uf = 0

½(m+M)v^2+1/2(I,pulley)w^2-mgh=0

½(m+M)v^2+1/2(1/2MR^2)(v^2/R^2)-mgh=0

so v = sqroot(2mgh/(M+m+1/2M,p))

v = sqroot(2(2kg)(9.81)(2.5)/(4kg+2kg+1/2(0.6kg))) =3.95m/s

b) w= v/r = (3.95m/s)/0.08m) = 49.5rad/s

I did solve it without the 0.25 coefficient of fiction though may you help me please?

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# Homework Help: Rotation with friction

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