Physics help?

Question

A cable is wrapped around a pulley in the shape of a solid cylinder(1/2mR^2) with rotational inertia I=0.250 kg-m^2 and radius R= 0.500m. The cylinder rotates without friction. The free end of the cable is connected to a block of mass m=1.0 kg on a rough inclined plane. This block is released from rest and slides down the incline to the bottom through vertical distance h=3.0m. During the motion, the plane exerts a frictional force on the block. At the bottom, the linear speed of the block is v=4.0 m/s.

a.) Use conservation of energy to compute the Heat (mu) that is generated during the motion.

please explain this one to me....

Answer 1

Look at the conserved energy of the system first at the bottom of the ramp:

The kinetic energy of the block
.5*m*v^2
or
.5*1*16=8 J

And the rotating energy of the pulley
.5*I*w^2
I was given as .250
Since the speed of the block translates through the cable, we know that w=v/r
so .5*.25*16/.25=8 J

The potential energy that was converted is
m*g*h
1*9.81*3
=29.43 J
so the frictional loss is
29.43-8-8=13.43 J

This, by the way is not mu. The mu*m*g*cos(th)*d equals the frictional work, which also
can be expressed as mu*m*g*3/tan(th) where th is the angle.
mu*1*3*/tan(th)=13.43
j