anonymous
  • anonymous
Hiii can someone help me with a Physics problem? Please and thank ya! :)
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anonymous
  • anonymous
oh boy, I looked at it! too late to get into though. you should get good help with that one, though it is tricky again..
anonymous
  • anonymous
OK, the work done is basically the force thats is influencing the movement times the distance, So we need to find the projection of the force in the direction of the movement, and it must be equal to the force of gravity summed with the friction, because it cannot have any aceleration for the speed to remain constant, and force=ma

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anonymous
  • anonymous
Considering that, we get:\[-mg-\mu N+Fsin \theta=0\]
anonymous
  • anonymous
The normal force is the force balancing the force thats pulling the block towards the wall so:\[N=Fcos \theta\]
anonymous
  • anonymous
So the work done is given by:\[W=\Delta xFsin \theta=(mg+\mu Fcos \theta)\Delta x\]
anonymous
  • anonymous
how do we find the change in x?
anonymous
  • anonymous
Now, whats missing is F. To find it we use the fact that the total force on the object must be zero. \[Fsin \theta=mg+\mu N=mg+\mu Fcos \theta\]Isolating F we get:\[Fsin \theta-F \mu \cos \theta =mg\]\[F=\frac{mg}{\sin \theta-\mu \cos \theta}\]Plug that into the formula for the work, and you have all the numbers now to calculate it. The change in x is given, it is 3m
anonymous
  • anonymous
so instead of N should i plug in Fcosθ
anonymous
  • anonymous
This is the work done by F, now to get the work done by gravity it is easy, we have the change in x and we also have the magnitude of the force, the only tricky part is that the work must be negative, because the force is going against the movement and not favoring it.
anonymous
  • anonymous
and what equation would we use for that?
anonymous
  • anonymous
Yes, for the first and\[W_G=-\Delta x mg\]for the second. Are you understanding what I'm doing here? Because this formula should be natural for you now.
anonymous
  • anonymous
a little, not really because i havent used it before
anonymous
  • anonymous
or at least i dont believe i have
anonymous
  • anonymous
It is just gravitational potential in disguise natasha! (GPE=mgh) the work done is just the potential energy 'gained' (but negative)
anonymous
  • anonymous
i understand the first part though, not really how you got the equation but how it makes sense
anonymous
  • anonymous
but then where is the h @furnessj
anonymous
  • anonymous
Yes, thats it, the h is x, just a diferent name.
anonymous
  • anonymous
h is the height gained, or just x here, anyway, leaving Ivan to it!
anonymous
  • anonymous
oh okay, so why cant we keep it as h?
anonymous
  • anonymous
oh okay!
anonymous
  • anonymous
If you understood everything so far, the only thing left is the normal force, wich you already have the formula for, and all the numbers involved. Just as a complementary comment for what furnessj said, the work done is also the change in the kinectic energy, and that is always true. However, the work is minus the change in the potential energy related to that force only if it is conservative, friction for example does work, but has no potential energy concept related to it.
anonymous
  • anonymous
Any doubts?
anonymous
  • anonymous
no, but how can i find the normal force formula?
anonymous
  • anonymous
@ivanmlerner ?
anonymous
  • anonymous
Oh, the normal force is the force that balances the force thats pulling the object towards the wall, so that it keeps still and don't "enter" the wall. It is the 3rd law of newton, that any force must have an opposite force balancing it. In this case what is pulling the object to the wall is the horizontal component of F, so its magnitude is Fcos theta
anonymous
  • anonymous
Oh, it would be that simple?
anonymous
  • anonymous
Yes, that simple
anonymous
  • anonymous
of course @natasha.aries , that simple :)
anonymous
  • anonymous
haha! okay thanks! can i try plugging it in a few minutes, and would you be able to tell me if its right?
anonymous
  • anonymous
ok... ur good learner .,
anonymous
  • anonymous
haha thank you!
anonymous
  • anonymous
To give a medal for asking do I just click on best response by the asker?
anonymous
  • anonymous
Yes!
anonymous
  • anonymous
and when i did the first part i got 204 does that sound about right?
anonymous
  • anonymous
I got approximately 312 using g=10.
anonymous
  • anonymous
the mass would be 5 correct?
anonymous
  • anonymous
yes
anonymous
  • anonymous
idk what i did wrong :/
anonymous
  • anonymous
Are you talking about part (a)?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Did you use a calculator? If you did, check for missing parentesis. Also, did you use w=xFsin theta? with that formula for F?
anonymous
  • anonymous
yes, it would be (5)(9.81) / (3sin30-.3cos30) right?
anonymous
  • anonymous
That would be F, but you got an extra 3 in the denominator in front of sin(30)
anonymous
  • anonymous
You still need to multiply that by 3, and by sin(30)
anonymous
  • anonymous
wait, im confused. so it would be (5)(9.81) / (sin30-.3cos30) but why do we need to multiply it by 3sin30 ?
anonymous
  • anonymous
anyone?
anonymous
  • anonymous
Sorry, I had to go in a rush, look at the formula for work on the begining of my posts:\[W=\Delta xFsin \theta\]So the one you are using is the one for F alone, and 3sin(30) is the rest: change in x=3 and sin theta=sin(30). Take a look again at the thinking I made to get to the first answer.
anonymous
  • anonymous
its all good! and now i get 306 :)
anonymous
  • anonymous
and so for the last one it would be 306(cos)30?
anonymous
  • anonymous
Using 9.81 as g that is correct, and for the second thats almost it.
anonymous
  • anonymous
306 is correct, but this is the work done by the force.
anonymous
  • anonymous
On the last it is asking for the normal, wich is a force, not work.
anonymous
  • anonymous
It is given by Fcos(30) and F is what you were calculating before.
anonymous
  • anonymous
F= that 204 you were calculating.
anonymous
  • anonymous
but Fcos(30) is the normal force?
anonymous
  • anonymous
i got 204 by mistake though
anonymous
  • anonymous
Yes, read what I wrote on the begining. You got 204 calculating the force alone, you had forgotten the other terms but here the force is exacly what you need, take a look at the begining and try to understand that, you don't seem to have understood it well.
anonymous
  • anonymous
i tried it again, and still didnt get that :/
anonymous
  • anonymous
im sorry :(
anonymous
  • anonymous
Wait, I said 204 is the force, not the answer, the answer is 204cos(30)
anonymous
  • anonymous
but i dont get 204 at all
anonymous
  • anonymous
The formulas are:\[N=Fcos \theta\]\[F=\frac{mg}{\sin \theta-\mu \cos \theta}\]
anonymous
  • anonymous
The explanation for them is in the begining.
anonymous
  • anonymous
i did that
anonymous
  • anonymous
Where you using that?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Show the numbers again.
anonymous
  • anonymous
5)(9.81) / (sin30-.3cos30)
anonymous
  • anonymous
And you need to multiply all of that by cos(30)
anonymous
  • anonymous
i finally got it! thats what ive been doing! i just separated it a few more times now. thank you so much. so the first one you would multiply it by 3sin30 and the last one cos30?
anonymous
  • anonymous
Yes, but remember what those numbers mean.
anonymous
  • anonymous
i will, im just unsure with how u knew which equations to use
anonymous
  • anonymous
I didn't I foun out wich one I needed to use, and I posted wverything I did, project the force to find the work, see what force is causing the normal, do the math and you get to the formulas.
anonymous
  • anonymous
when i checked the answers, the right answers were 310J... -150J...180 N
anonymous
  • anonymous
310 for a -150 for b and 180 for c
anonymous
  • anonymous
i got a similar answer for the first 2 parts, but not for the third!
anonymous
  • anonymous
Remember that on the last one you only need to multiply the force F (204N) by cos(30) (sqrt(3)/2) because we need the projection of the force at the direction perpendicular to the wall. If you use those numbers you'll get the answer.
anonymous
  • anonymous
why sqrt of 3/2 ?
anonymous
  • anonymous
i think you just mean cos30 because those two are the same things :)
anonymous
  • anonymous
What?? wait, thats what i meant:\[\cos 30°=\frac{\sqrt{3}}{2}\]

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