Two buckets of mass m1=19.9 kg and m2=13.7 kg are attached to the ends of a massless rope which passes over a pulley with a mass of mp=8.73 kg and a radius of =0.150 m. assume that the rope does not slip on the pulley, and that the pulley rotates without friction. the buckets are released from rest and begin to move. if the larger bucket is a distance =1.75 m above the ground when it is released, with what speed will it hit the ground?

This question involves the concepts of the law of conservation of energy, potential energy, and kinetic energy.

The speed with which the larger bucket hits the ground will be "2.24 m/s".

Applying the law of conservation of energy to this situation we have:

Change in Potential Energy = Change in Kinetic Energy

where,

m₁ = mass of larger bucket = 19.9 kg

m₂ = mass of smaller bucket = 13.7 kg

h = height = 1.75 m

v = speed of buckets = ?

r = radius of pulley = 0.15 m

mp = mass of pulley = 8.73 kg

I = moment of inertia of pulley =

ω = angular speed =

Therefore,

v = 2.24 m/s

Learn more about the law of conservation of energy here:

link

The attached picture explains the law of conservation of energy.

Explanation:

As we know that here larger bucket will move down and smaller bucket will move up by same distance

So by energy conservation we can say that decrease potential energy = gain in kinetic energy of the system

So we will have

also we know that string moves over the pulley without slipping

so we will have

now we have

1/0.14nm = 10^10/1.4 a lot