GPE and KE of two unequal masses over pulley

[ethex]

Homework Statement

Two unequal masses are connected by a light cord passing over a smooth pulley. State the gravitational potential energy \DeltaU(> or < or = 0) and the kinetic energy \DeltaK(> or < or = 0) of the system after the masses are released from rest?

Homework Equations

The Attempt at a Solution

Both masses were originally at the same level. After released, both went in an opposite direction away from the equilibrium level. Thus, \DeltaU=0?

Both gained speed, thus \DeltaK>0?

 

[grzz]

Note that potential evergy is a scalar and hence has not direction.

 

[ethex]

Hi grzz, So there is a difference in height, thus there is a change in GPE, How do i determine if change in GPE is >0 or <0?

 

[grzz]

To use grav p e = mgh one needs a zero reference level. To make things simple you can choose the lowest position reached as this level. A diagram always helps.

 

[ethex]

I am confused. So which is the initial height and final since there are two masses?

 

[grzz]

You have to assume the initial (equal) height, say h. Continue from there.

 

[ethex]

For heavier mass, \DeltaU = mg(0-h) = -mgh
For lighter mass, \DeltaU = mg (2h- h) = mgh

Thus for the whole system = 0?

 

[grzz]

What height are you taking as the reference for GPE = 0?

 

[ethex]

Lowest point reached as you have suggested.

 

[grzz]

Your only mistake is that the masses are not equal. Hence use m and M.

 

[grzz]

Any way. The bigger one goes down,hence U decreased and the other goes ...

 

[ethex]

That's right! I missed out the difference in masses.

For heavier mass, change in U = -Mgh
For lighter mass, change in U = mgh

Since M>m, total change in U < 0.

As for KE, as long as there is a movement, does it mean KE is affected?

So the ans to this question is change in u< 0 , change in K>0, right?

And is there any situation change in KE<0?

I am quite confused over all these.

Thank you so much grzz.

 

[grzz]

initial KE = 0 since they were at rest.
Hence KE increased for both.

 

[ethex]

I got it, thank you.

Would you mind helping me out another question relating to work energy?

A 3 kg ball swings rapidly in a complete vertical circle of radius 2 m by a light string. The ball moves so fast that the string is always taut. As the ball swings from its lowest point to its highest point,

the work done on it by gravity is ___ and the work done on it by the tension in the string is ___.

The first blank i found out to be -188J. How do it work on the second part?

 

[grzz]

Can you show the working for -188J?

 

[ethex]

Ok.

Change in GPE = mgh = 3 x 4 x 9.8 = 188J
Since work done is by the gravity on the system, therefore -188J.

 

[grzz]

-118j

 

[grzz]

Now can you state the definition of work?

 

[grzz]

The CORRECT defn of work will give you the answer for the work by the tension.

 

[ethex]

Work done is the product of force that acts on the object with the displacement along the action of force.

The top tension is pointing down while tension at bottom is pointing up with larger magnitude. So how do i cal the work done by the tension base on definition?

 

[grzz]

Work done is the product of force that acts on the object with the displacement along the action of force.

Correct defn. Note the phrase 'dislplacement along the force (i.e. along the tension).
Now look at the displacement all along the circle. Is the displacement ever along the tension?

 

[ethex]

Since the motion is in a circular path and the tension is always pointing towards the centre, thus work done by the tension is zero?

But if that make sense, then wouldn't the first part be zero as well? There is no displacement along the gravity?

 

[grzz]

Work by tension is 0. CORRECT!
BUT when moving along the circle, the displacement WILL have a component parallel to gavity i.e. upwards or downwards except at two particular points.

 

[ethex]

Oh i Got it! Thank you grzz and have a great day ahead!

 

[grzz]

You are welcome. Isn't Physics interesting!