Can I use the principle of conservation of energy for this problem?

[hendrix7]

Homework Statement

I have attached an image of the question.
I am trying to tackle part (d) and thought I could use conservation of energy but my numbers don't come out right.

Relevant Equations

The potential energy lost by P in travelling to the ground is 4mg x h sin alpha (3/5) = 12mgh/5
The work done by P against friction is (4mg x cos alpha (4/5) x h)/4 = 4mgh/5
Therefore, the gain in potential energy of Q is 8mgh/5 and this would give an increase in height of Q of 8h/5, but the question simply states that this should be at least 28h/25 so can someone put me straight as to what I'm missing?

ddd

 

[kuruman]

Homework Statement:: I have attached an image of the question.
I am trying to tackle part (d) and thought I could use conservation of energy but my numbers don't come out right.
Relevant Equations:: The potential energy lost by P in traveling to the ground is 4mg x h sin alpha (3/5) = 12mgh/5
The work done by P against friction is (4mg x cos alpha (4/5) x h)/4 = 4mgh/5
Therefore, the gain in potential energy of Q is 8mgh/5 and this would give an increase in height of Q of 8h/5, but the question simply states that this should be at least 28h/25 so can someone put me straight as to what I'm missing?

View attachment 285256ddd

Please show the equation that you used to get your answer. With that equation missing from your post, we cannot figure out what you missed in the equation.

 

[Charles Link]

I think the missing part is that P has kinetic energy as it hits the ground that is simply lost, and this energy needs to be computed and subtracted from the available energy for Q .

 

[kuruman]

I think the missing part is that P has kinetic energy as it hits the ground that is simply lost, and this energy needs to be computed and subtracted from the available energy for Q .

Yes, in which case Q keeps moving up until it barely reaches the pulley.

On edit: I get a value somewhat higher than $\dfrac{28h}{25}.$

 

[Charles Link]

I did get the same 8/5 factor that the OP got, before I realized this kinetic energy that P has needs to get subtracted out.

 

[Charles Link]

On edit: I get a value somewhat higher than

The available kinetic energy after traveling a distance $h$ gets split between them, with the amount each gets, (each with the same $v$), is proportional to the mass. The kinetic energy that gets split between them is the potential energy of P, minus the friction energy, and minus the gain of potential energy of mass Q in rising a distance $h$.

The next step, once the kinetic energy of P is computed, is to set $mgd>$ Energy available from P.

I agree with the d> 28h/25 answer.

alternatively, one could compute the kinetic energy of Q after the distance $h$, and set this equal to $mg \Delta$ , and then we must have $d> h+\Delta$.

 

[kuruman]

alternatively, one could compute the kinetic energy of Q after the distance $h$, and set this equal to $mg \Delta$ , and then we must have $d> h+\Delta$.

That's how I did it at first and got $d>\dfrac{31h}{25}$. After redoing it, more carefully, I got $d>\dfrac{28h}{25}.$ All is well.

 

[hendrix7]

I think the missing part is that P has kinetic energy as it hits the ground that is simply lost, and this energy needs to be computed and subtracted from the available energy for Q .

Of course! Thank you so much. How did I miss that?