How Does the Incline Plane Affect Potential Energy in a Two-Mass System?

[Larrytsai]

If I have a mass m1 sitting on an incline plane attached by a rope and a pulley to another mass m2 hanging off the incline plane, what would be the potential energy?

I was thinking that m1 only plays a role on the potential energy.

 

[SammyS]

If I have a mass m1 sitting on an incline plane attached by a rope and a pulley to another mass m2 hanging off the incline plane, what would be the potential energy?

I was thinking that m1 only plays a role on the potential energy.

Can m₂ also rise and/or fall ?

 

[Larrytsai]

m2 can rise or fall

 

[SammyS]

m2 can rise or fall

In that case, its potential energy changes.

Is m₂ part of the system ?

 

[Larrytsai]

yes m2 is conneced by a rope and I would like to find the velocity of m1 after it leaves the ramp.

So if it does change the systems potential energy would it just be

PE = g*(m1*height + m2*height)

the height of each respective mass are different.

 

[SammyS]

yes m2 is connected by a rope and I would like to find the velocity of m1 after it leaves the ramp.

So if it does change the systems potential energy would it just be

PE = g*(m1*height + m2*height)

the height of each respective mass are different.

Yes, the height of each respective mass is different. All that matters is change in height. Mass, m₁, is on an inclined plane, so when the height of m₂ decreases by Δh, the height of m₁ increases by a lesser amount. Use trig to relate the two.

 

[QuantumJock]

Potential energy will not affected by the other mass.

PE = mgh , mass x gravity x height.

You know the mass, and you know gravity is equal to 9.81 so you will need to know the height.

Jock

 

[SammyS]

Potential energy will not affected by the other mass.

PE = mgh , mass x gravity x height.

You know the mass, and you know gravity is equal to 9.81 so you will need to know the height.

Jock

Hello QuantumJock. Welcome to PF !

You have replied to a thread which has been idle for several months .

 

[QuantumJock]

Oh woops I'm sorry! Still new!

Jock