Finding Velocity Using Work-Energy Principle

[Bones]

Homework Statement

The system of two blocks is released from rest. The pulley is a solid cylinder of mass 5.00kg and radius of 10.0cm. MA is 10.0kg and MB is 20.0kg. The coefficient of kinetic friction between MA and the table is 0.2. Find the velocity of the blocks after moving 3.00m using work-energy principle.

Block A is on a flat surface and block B is hanging. They are connected by a rope over a pulley.

Homework Equations

The Attempt at a Solution

I don't remember how to do this, please help!

 

[tiny-tim]

Welcome to PF!

The system of two blocks is released from rest. The pulley is a solid cylinder of mass 5.00kg and radius of 10.0cm. MA is 10.0kg and MB is 20.0kg. The coefficient of kinetic friction between MA and the table is 0.2. Find the velocity of the blocks after moving 3.00m using work-energy principle.

Block A is on a flat surface and block B is hanging. They are connected by a rope over a pulley.
…
I don't remember how to do this, please help!

Hi Bones! Welcome to PF!

Calculate the work done by the friction force on MA and by gravity on MB.

(and remember that the two blocks, and the rope on the pulley, will have the same speed!)

That will equal the change in KE.

 

[Bones]

Work Force of friction: 0.2*98N*(d)(cos180)=-19.6(d)
Change in KE=1/2(20kg)(V)^2-1/2(20kg)(0)

-19.6(d)=1/2(20kg)(V)^2

I don't think I am doing this right...

 

[tiny-tim]

Work Force of friction: 0.2*98N*(d)(cos180)=-19.6(d)
Change in KE=1/2(20kg)(V)^2-1/2(20kg)(0)

-19.6(d)=1/2(20kg)(V)^2

What about gravity? and KE of the pulley?

 

[Bones]

mAgh+0.5mAvA^2=mBgh+0.5mBvB^2

(0.2)(98N)(3.00m)(cos180)=(10kg)(9.8m/s^2)(0)+(0.5)(10kg)(vA)^2=(20kg)(9.8m/s^2)(h)+(0.5)(20kg)(vB)^2

I am not sure about the height of mass B. I don't think this is right either...

 

[Bones]

I think I figured it out. Thanks for your help!