# Calculating the acceleration and kinetic energy of a bucket in a pulley system

1. The problem statement, all variables and given/known data

A well pulley is constructed from wooden cylinder with a mass of 5 kg and a radius of 10 cm and a light handle. The bucket filled with water is hanging on the rope coiled to the cylinder has a mass of 7 kg.
- With how much acceleration does the bucket move if the handle breaks; the pulley is spinning with no friction?
- How much is the kinetic energy of the bucket after it falls for 1 m?

2. Relevant equations

F=ma
KE= mv²/2
PE= mgh

3. The attempt at a solution

First part: With how much acceleration does the bucket move if the handle breaks; the pulley is spinning with no friction?

m(1)= 5 kg, m(2)= 7 kg, g= 10 m/s²

(m(1) + m(2))*a= m(2)g
a= m(2)g / (m(1) + m(2)
a= 5.8 m/s²

ARE MY CALCULATIONS CORRECT?

Second part: How much is the kinetic energy of the bucket after it falls for 1 m?

KE= PE
mv²/2= mgh
v= sqrt(2gh)
v= 4.47 m/s

KE= mv²/2= 69.9 J

Are my calculations correct?

Thank you helping!

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#### rl.bhat

Homework Helper
The equation for acceleration is wrong. Check it.

Do I have to take into the consideration inertia and the torque of the pulley (I know the radius and the mass of the pulley) and develop it from there?

What about the second part? Is it correct?

Thank you for helping!

Last edited:

#### Jebus_Chris

There is torque acting on the cylinder because of the tension in the rope.

Please, can you give a hint how to apply the torque into the calculations? I'm really confused!?

I tried it like this:

For the bucket:
m(1)a= m(1)g - T

For the pulley:
Iα= Tr

I= 1/2m(2)r² and α= a/r

1/2m(2)r²*a/r = Tr → 1/2m(2)r= T

For the system:
1/2m(2)a + m(1)a = m(1)g -T + T
a= m(1)g / (m(1) + 1/2m(2))
a= 7 m/s²

I really need help with solving this problem. Thank you for helping!

#### Jebus_Chris

Yes.

Now for the second part you just replace g with your new a.

Thank you very much!

So the correct answer is: 49 J

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