# A mass m1 = 15.0 kg and a mass m2 = 11.5 kg are suspended by a pulley that has a radius of 10.0 cm and a mass of 3.00 kg (Fig. P10.43). The?

A mass m1 = 15.0 kg and a mass m2 = 11.5 kg are suspended by a pulley that has a radius of 10.0 cm and a mass of 3.00 kg (Fig. P10.43). The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest 3.00 m apart. Treating the pulley as a uniform disk, determine the speeds of the two masses as they pass each other.

answer and process please, thanks!

### 2 Answers

- NCSLv 72 months agoFavourite answer
You could use either energy or kinematics here -- I think they want the former.

When they pass each other, they will each have moved 1.50 m and the decrease in potential energy is

PE = (15.0 - 11.5)kg * 9.81m/s² * 1.50m = 51.5 J

This becomes KE:

KE = ½(m1 + m2)v² + ½Iω²

and assuming a uniform disk,

I = ½Mr²

and because the rope doesn't slip, ω = v/r. That makes

½Iω² = ½*½Mr²*(v/r)² = ¼Mv²

and it turns out that the radius is immaterial.

Then 51.5 J = ½*26.5kg*v² + ¼*3.00kg*v²

which solves to v = 1.92 m/s ◄

Hope this helps!

- oubaasLv 72 months ago
m1 = 15 kg

m2 = 11.5 kg

Me = M/2 = 1.5 kg

acceleration a = g(m1-m2)/(m1+m2+Me) = 9.806*3.5/28 = 1.226 m/sec^2

V = √ 2*a*h/2 = √ 3*1.226 = 1.92 m/sec ....in just two lines 😉