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anonymous
 3 years ago
Hiii can someone help me with a Physics problem? Please and thank ya! :)
anonymous
 3 years ago
Hiii can someone help me with a Physics problem? Please and thank ya! :)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh 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
 3 years ago
Best ResponseYou've already chosen the best response.0OK, 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Considering that, we get:\[mg\mu N+Fsin \theta=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The normal force is the force balancing the force thats pulling the block towards the wall so:\[N=Fcos \theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So the work done is given by:\[W=\Delta xFsin \theta=(mg+\mu Fcos \theta)\Delta x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how do we find the change in x?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, 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:\[Fsin \thetaF \mu \cos \theta =mg\]\[F=\frac{mg}{\sin \theta\mu \cos \theta}\]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
 3 years ago
Best ResponseYou've already chosen the best response.0so instead of N should i plug in Fcosθ

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This 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
 3 years ago
Best ResponseYou've already chosen the best response.0and what equation would we use for that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.0a little, not really because i havent used it before

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or at least i dont believe i have

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It is just gravitational potential in disguise natasha! (GPE=mgh) the work done is just the potential energy 'gained' (but negative)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i understand the first part though, not really how you got the equation but how it makes sense

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but then where is the h @furnessj

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, thats it, the h is x, just a diferent name.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0h is the height gained, or just x here, anyway, leaving Ivan to it!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh okay, so why cant we keep it as h?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If 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
 3 years ago
Best ResponseYou've already chosen the best response.0no, but how can i find the normal force formula?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, 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
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, it would be that simple?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0of course @natasha.aries , that simple :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha! okay thanks! can i try plugging it in a few minutes, and would you be able to tell me if its right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok... ur good learner .,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To give a medal for asking do I just click on best response by the asker?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and when i did the first part i got 204 does that sound about right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got approximately 312 using g=10.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the mass would be 5 correct?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0idk what i did wrong :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Are you talking about part (a)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did you use a calculator? If you did, check for missing parentesis. Also, did you use w=xFsin theta? with that formula for F?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, it would be (5)(9.81) / (3sin30.3cos30) right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That would be F, but you got an extra 3 in the denominator in front of sin(30)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You still need to multiply that by 3, and by sin(30)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait, im confused. so it would be (5)(9.81) / (sin30.3cos30) but why do we need to multiply it by 3sin30 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, 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
 3 years ago
Best ResponseYou've already chosen the best response.0its all good! and now i get 306 :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and so for the last one it would be 306(cos)30?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Using 9.81 as g that is correct, and for the second thats almost it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0306 is correct, but this is the work done by the force.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0On the last it is asking for the normal, wich is a force, not work.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It is given by Fcos(30) and F is what you were calculating before.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0F= that 204 you were calculating.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but Fcos(30) is the normal force?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i got 204 by mistake though

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.0i tried it again, and still didnt get that :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait, I said 204 is the force, not the answer, the answer is 204cos(30)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but i dont get 204 at all

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The formulas are:\[N=Fcos \theta\]\[F=\frac{mg}{\sin \theta\mu \cos \theta}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The explanation for them is in the begining.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Where you using that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Show the numbers again.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.05)(9.81) / (sin30.3cos30)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And you need to multiply all of that by cos(30)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, but remember what those numbers mean.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i will, im just unsure with how u knew which equations to use

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0when i checked the answers, the right answers were 310J... 150J...180 N

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0310 for a 150 for b and 180 for c

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i got a similar answer for the first 2 parts, but not for the third!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Remember 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
 3 years ago
Best ResponseYou've already chosen the best response.0i think you just mean cos30 because those two are the same things :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What?? wait, thats what i meant:\[\cos 30°=\frac{\sqrt{3}}{2}\]
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