Answer
The linear speed of a point on belt 2 is $~~0.75~m/s$
Work Step by Step
In part (c), we found that the angular speed of pulley B' is $~~15~rad/s$
We can find the linear speed of a point on belt 2:
$v = \omega_{B'}~R_{B'}$
$v = (15~rad/s)(0.05~m)$
$v = 0.75~m/s$
The linear speed of a point on belt 2 is $~~0.75~m/s$
