Rotation with friction

[cristina]

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?


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?

[NateTG]

How much work does the friction do?

[cristina]

49.5rad/s * 0.25?

[NateTG]

No.

Work = Force X Distance

It will be an extra term in the energy equation.

[NateTG]

The work done by friction is going to be force times distance.

This would be easier to illustrate with a drawing, but it's going to be:
F_f=mg\mu_k
so
F_f=4 (9.81) 0.25=9.81
mow
W_f=9.81*2.5
you can plug that into your energy equation:
\frac{1}{2}(m_1+m_2)v^2+\frac{1}{2}I\omega^2+W_f-mgh=0

I think you've got the math under control from there.

[cristina]

thank you very much, I got it.

[NateTG]

In response to your post:

E_{final}=E_{inital}

The work done by friction typically ends up as a temeperature change on the E_{final} side. It's just one more thing that you need to account for in the energy equation