Pulley Problem on an Inclined Surface Using Energy Methods

[Tygra]

Homework Statement

How far does particle A trave after string snaps

Relevant Equations

See in post

Dear all I have the following problem with my working attached as a PDF.

I want to calculate how far particle a travels after the string has snapped. The question says the particle A travel for 1m before the string snaps. The question asks me to use energy methods to do this. The total distance is 1.51m according to the book. I am getting 1.52 m, but it my working I am trying to solve for the extra distance after the particle has travelled the 1m. Thus, I am looking to get 0.51m.

Please see my working in the attachment.

Many thanks

 

[erobz]

I assume they are supposed to be "blocks" not objects that can roll? $F$ is a kinetic friction force, and mass $B$ was descending (mass a climbing the incline) before it breaks? How do you get that $T = 5 mg$? The clarity of your problem statement leaves a bit to be desired with missing/unexplained parameters.

 

[Tygra]

I assume they are supposed to be "blocks" not objects that can roll?

I guess so. This is the way the diagram is illustrated in the book.

F is a kinetic friction force, and mass B was descending (mass a climbing the incline) before it breaks?

Yes

How do you get that T=5mg?

It's the tension in the string, obtained by resolving forces in the vertical considering particle B.

T - 5mg = 0 T = 5mg

The clarity of your problem statement leaves a bit to be desired with missing/unexplained parameters.

This is the question from the book:

The diagram shows a particle A of mass 2m which can move on a rough surface of a plane inclined at an angle theta to the horizontal, where sin(theta) = 3/5. A second particle B of mass 5m hangs freely attached to a light inextensible string which passes over a small pulley. The other end of the string is attached to particle A. The coefficient of friction between particle A and the plane is 3/8. particle B is initially hanging 2m above the ground and particle A is 4m from the pulley. When the system is released from rest with the string taught A moves up a line of greatest slope of the planes.

a) Find the initial acceleration of A? (answered)

When B has descended 1m the string breaks
b) By using the principle of conservation of energy calculate the total distance moved by A before it comes to rest?

 

[erobz]

T - 5mg = 0 T = 5mg

Did you consider the acceleration of particle $B$?

 

[haruspex]

Did you consider the acceleration of particle $B$?

To put that more pointedly, @Tygra, if the tension is equal to the weight of B, how will B move?

 

[kuruman]

In addition to @erobz's and @haruspex's remarks, I point out that

The coefficient of friction between particle A and the plane is 3/8.

Your solution completely ignores friction. You need to provide a revised version incorporating our comments. Please do not provide pdf attachments. We prefer that you use $\LaTeX$ and post online. To learn how, click the link "LaTeX Guide". lower left above "Attach file". Provisional photos of your work as attached files may be acceptable for a short time as long as they are easy to read.

 

[Tygra]

In addition to @erobz's and @haruspex's remarks, I point out that

Your solution completely ignores friction.

No it does not. I have calculated the frictional force as 24/40mg

You need to provide a revised version incorporating our comments. Please do not provide pdf attachments. We prefer that you use $\LaTeX$ and post online. To learn how, click the link "LaTeX Guide". lower left above "Attach file". Provisional photos of your work as attached files may be acceptable for a short time as long as they are easy to read.

Okay, sorry about that. I know not to do it again

 

[Tygra]

To put that more pointedly, @Tygra, if the tension is equal to the weight of B, how will B move?

Yes true, lol, silly me!

 

[Steve4Physics]

Hi @Tygra. Despite mistakes I’d say your general strategy is OK.

No one has yet mentioned the following; you have written:
$(\frac 65mg + \frac {24}{40}mg)x = 8.96m + \frac 65mgx$
which you need to revisit.

Make sure you understand the physical meaning of the terms in the equation and make sure there is no ‘double-counting’ of any sort. (It may help to remember that work done by gravity is the same thing as minus the change in gravitational potential energy.)

Correcting and solving the above equation gives the required answer (fortunately, despite any other mistakes you have made)!

You will find it a very useful learning-exercise to present a complete, updated solution here, as already suggested, and to get the feedback.

 

[kuruman]

No it does not. I have calculated the frictional force as 24/40mg

Yes, thank you. I see now what you did. You considered the two-mass system with the masses "straightened out." Then the net force on the system is $$F_{\text{net}}=m_2g-m_1g\sin\theta-\mu m_1g\cos\theta.$$ Then you applied Newton's second law to find the acceleration of the two-mass system, $$a= \frac{F_{\text{net}}}{m_2+m_1}=\frac{m_2g-m_1g\sin\theta-\mu m_1g\cos\theta}{m_2+m_1}=
\frac{5m-2m(\frac{3}{5})-(\frac{3}{8}) 2m(\frac{4}{5})}{7m}g=\frac{16}{35}g.$$Note that this gives the same numerical answer that you got. Having it in symbolic form and substituting numbers at the very end makes it easier to read and follow. Keeping it in fractional format avoids round-off errors. Also note that since you picked both masses as your system, the tension is an internal force and its value does not matter because it does not enter any equation.

 

[Tygra]

Hi @Tygra. Despite mistakes I’d say your general strategy is OK.

No one has yet mentioned the following; you have written:
$(\frac 65mg + \frac {24}{40}mg)x = 8.96m + \frac 65mgx$
which you need to revisit.

Hi Steve,

Yes I knew beforehand that this would the underlying issue to the problem.

Here was my thinking:

The kinetic energy at the moment before the string breaks is 8.96 Joules. Since the particle A comes to rest then change in kinetic energy also equal 8.96 joules.

The gravitational potential energy increases. I have taken the datum level for the GPE to be the height the moment the string breaks (Not sure if this is correct - probably not).

So the GPE gained in 2mg*sin(theta)x = 2mg*3/5*x = 6/5mgx (where x is the distance particle a travels up the plane before coming to rest).

Then by considering the energy change you get 8.96m joules + 6/5mgx joules. Then equate this to the work done on particle A.

I'm not quite sure where to go from here. So any more help would be much appreciated. Thanks

 

[kuruman]

You have the Joules of initial kinetic energy. Do all of these Joules go into gravitational potential energy increase?

 

[Steve4Physics]

Since the particle A comes to rest then change in kinetic energy also equal 8.96 joules.

KE is reduced, so the change (ΔKE) is in fact -8.96J.

(If you work symbolically (preferred), this is −3235g EDIT: whoops, that should be $-\frac {32}{35}mg$.)

The gravitational potential energy increases. I have taken the datum level for the GPE to be the height the moment the string breaks (Not sure if this is correct - probably not).

So the GPE gained in 2mg*sin(theta)x = 2mg*3/5*x = 6/5mgx (where x is the distance particle a travels up the plane before coming to rest).

Then by considering the energy change you get 8.96m joules + 6/5mgx joules. Then equate this to the work done on particle A.

You are incorrectly accounting the change in GPE.

W = $\frac 65mg$ is the downhill component of A's weight. The work done by W is the same thing as {minus the change in A's GPE).

The work-energy theorem essentially says:
Total work done by external force(s) = ΔKE

Here, the left side of the above equation can be expressed as:
a) Work done F + Work done by W
or
b) Work done F + (-gain in A's gravitational potential energy)

a) and b) mean exactly the same thing!

Have another go!

Editted.

 

[Tygra]

Thanks Steve and Kuruman, I have the correct answer now.

Like you said, Steve, I can set it up like this:

$(\frac {24mg}{40} + \frac{6mg}{5})x = \frac{224m}{25}$

$\frac {9mg}{5} x = \frac{224m}{25}$

$x = \frac{225(5)}{25(9)(9.8)} = 0.51m$

Thank you again for your help!

 

[Steve4Physics]

$(\frac {24mg}{40} + \frac{6mg}{5})x = \frac{224m}{25}$

Well done! Note that the right side (which corresponds to $-\Delta (KE)$) is actually $\frac {32}{35}mg$; that means $mg$ cancels-out from all terms of the equation giving:
$(\frac {24}{40} + \frac{6}{5})x = \frac{32}{35}$

Working symbolically has revealed the interesting fact that the answer doesn't depend on the actual values of $m$ or $g$. E.g. you get the same answer if the experiment is conducted on earth or on the moon!

Edit: typo's fixed.

 

[Tygra]

Well done! Note that the right side (which corresponds to $-\Delta (KE)$) is actually $\frac {32}{35}mg$; that means $mg$ cancels-out from all terms of the equation giving:
$(\frac {24}{40} + \frac{6}{5})x = \frac{32}{35}$

Thanks, Steve

Working symbolically has revealed the interesting fact that the answer doesn't depend on the actual values of $m$ or $g$. E.g. you get the same answer if the experiment is conducted on earth or on the moon!

Edit: typo's fixed.

That is a very interesting fact. I wouldn't have sussed it unless you said.

 

[erobz]

As an alternative you can start with:

$$ \int \mathbf{F_{n.c.}} \cdot d \mathbf{\ell} = \Delta KE + \Delta PE \tag{eq1} $$

$$ -2\mu m g \cos \theta x = - 5m g x + 2m g \sin \theta x + \frac{1}{2}( 2m + 5m ) v^2 \tag{eq2}$$

Then you can solve directly for $v^2$ (the initial velocity of the masses just before the rope snaps). Then you can re-apply $\text{eq1}$ to the mass on the incline to solve part b)

to get the acceleration $a$ from part a) just differentiate $\text{eq2}$ w.r.t. time $t$.