## Machines B/C

Test your knowledge of various Science Olympiad events.
Godspeed
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### Re: Machines B/C

RiverWalker88 wrote:
September 29th, 2020, 8:51 pm
astronomybuff wrote:
September 29th, 2020, 5:45 pm
RiverWalker88 wrote:
September 29th, 2020, 4:59 pm

Thanks, Umaroth for pointing out that you can't calculate the AMA of the system, thus making parts c and d impossible.
Wow!! So it wasnt just me thank God. I was kinda scared to ask ngl.
Nope, It wasn't! When in doubt, I probably wrote the question wrong...
Yeah I thought I was just dumb and didn't know the AMA. If not specified tho, isn't the AMA always the same as the IMA if you're not given any extra information such as friction?

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### Re: Machines B/C

Godspeed wrote:
September 29th, 2020, 10:17 pm
RiverWalker88 wrote:
September 29th, 2020, 8:51 pm
astronomybuff wrote:
September 29th, 2020, 5:45 pm

Wow!! So it wasnt just me thank God. I was kinda scared to ask ngl.
Nope, It wasn't! When in doubt, I probably wrote the question wrong...
Yeah I thought I was just dumb and didn't know the AMA. If not specified tho, isn't the AMA always the same as the IMA if you're not given any extra information such as friction?
Generally speaking, yes, you will assume IMA and AMA are the same for a system, especially if the question states that the system is ideal. However, if you specifically are asked for AMA, you will probably have to find some way to get the effort and resistance forces to do this. I gave no effort force for this, so it was impossible to solve for the AMA.
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Godspeed
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### Re: Machines B/C

RiverWalker88 wrote:
September 29th, 2020, 10:22 pm
Godspeed wrote:
September 29th, 2020, 10:17 pm
RiverWalker88 wrote:
September 29th, 2020, 8:51 pm

Nope, It wasn't! When in doubt, I probably wrote the question wrong...
Yeah I thought I was just dumb and didn't know the AMA. If not specified tho, isn't the AMA always the same as the IMA if you're not given any extra information such as friction?
Generally speaking, yes, you will assume IMA and AMA are the same for a system, especially if the question states that the system is ideal. However, if you specifically are asked for AMA, you will probably have to find some way to get the effort and resistance forces to do this. I gave no effort force for this, so it was impossible to solve for the AMA.
Oh I see, ok.

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### Re: Machines B/C

Is e 5827.14 joules?

RiverWalker88
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### Re: Machines B/C

So, for the scattered answers (thanks for posting, this was a very poorly-written question, sorry ) here's the solutions.
1. Block & Tackle
2. IMA = 4. There are two moving pulleys, so the IMA is 2*2=4
3. This is why I should solve
4. problems before posting them.
5. I got 1903 J, because I used the frictional force that needs to be overcome for the force that the block was pulled at. Once again, though, this part was unclear and far too open to assumption...

So, yeah. Someone's turn now. Next time I post a question, I will solve it first to ensure clarity and possibility. Apologies for the bad question.
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astronomybuff
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### Re: Machines B/C

Alright, it's been a while.

A block slides down a plane inclined at an angle of θ degrees with constant velocity. It is then projected up the same plane.

Determine the following (symbolically)
a) Coefficient of kinetic friction
b) The distance it travels before it stops
c) Whether or not it will slide down again

astronomybuff
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### Re: Machines B/C

Hello....
Don't leave me hanging guys
Last edited by astronomybuff on October 10th, 2020, 2:42 pm, edited 1 time in total.

Hiujier
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### Re: Machines B/C

a)
Since the block slides down parallel to the plane at a constant velocity, we know that the net force parallel to the plane is 0. Drawing a free body diagram and writing Newton's 2nd Law, we get:

$mg \sin \theta - f = 0$
$mg \sin \theta - \mu_k N = 0$
$mg \sin \theta - \mu_k mg \cos \theta = 0$
$mg \sin \theta = \mu_k mg \cos \theta$
$\mu_k = \frac{\sin \theta}{\cos \theta}$
b)
I'm not sure if you can find the distance without initial velocity at the bottom of the ramp, so here is my solution given intial velocity:

Using energy conservation with work done by friction, we get:

$\frac{1}{2} mv^2 = fd + mgd \sin \theta$,

where $d$ is the distance traveled by the block parallel to the plane.

Furthermore, we know from part (a) that:
$f = mg \sin \theta$, so:

$\frac{1}{2} mv^2 = mgd \sin \theta + mgd \sin \theta$

$\frac{1}{2} mv^2 = 2mgd \sin \theta$
$d = \frac{v^2}{4g \sin \theta}$
c)
When the block is at rest after sliding, we know it experiences static friction. For static friction, we know that:

$\mu_s \geq \mu_k$

From this, we know that:
$f_{static} \geq f_{kinetic} = mg \sin \theta$,
$f_{static} \geq mg \sin \theta$

So the combining both conditions we get that the net force must be 0 when the block is at rest at the top of the plane. Therefore, it will not slide down again.
Last edited by Hiujier on October 10th, 2020, 8:01 pm, edited 2 times in total.
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astronomybuff
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### Re: Machines B/C

Yep! Everything correct. Yes, my bad I forgot to include the velocity with which it was projected is v0. Your turn!

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