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

- anonymous

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

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- anonymous

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- 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

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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## More answers

- anonymous

Considering that, we get:\[-mg-\mu N+Fsin \theta=0\]

- anonymous

The normal force is the force balancing the force thats pulling the block towards the wall so:\[N=Fcos \theta\]

- anonymous

So the work done is given by:\[W=\Delta xFsin \theta=(mg+\mu Fcos \theta)\Delta x\]

- anonymous

how do we find the change in x?

- 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

so instead of N should i plug in Fcosθ

- 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

and what equation would we use for that?

- 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

a little, not really because i havent used it before

- anonymous

or at least i dont believe i have

- anonymous

It is just gravitational potential in disguise natasha! (GPE=mgh) the work done is just the potential energy 'gained' (but negative)

- anonymous

i understand the first part though, not really how you got the equation but how it makes sense

- anonymous

but then where is the h @furnessj

- anonymous

Yes, thats it, the h is x, just a diferent name.

- anonymous

h is the height gained, or just x here, anyway, leaving Ivan to it!

- anonymous

oh okay, so why cant we keep it as h?

- anonymous

oh okay!

- 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

Any doubts?

- anonymous

no, but how can i find the normal force formula?

- anonymous

@ivanmlerner ?

- 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

Oh, it would be that simple?

- anonymous

Yes, that simple

- anonymous

of course @natasha.aries , that simple :)

- 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

ok... ur good learner .,

- anonymous

haha thank you!

- anonymous

To give a medal for asking do I just click on best response by the asker?

- anonymous

Yes!

- anonymous

and when i did the first part i got 204 does that sound about right?

- anonymous

I got approximately 312 using g=10.

- anonymous

the mass would be 5 correct?

- anonymous

yes

- anonymous

idk what i did wrong :/

- anonymous

Are you talking about part (a)?

- anonymous

yes

- 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

yes, it would be (5)(9.81) / (3sin30-.3cos30) right?

- anonymous

That would be F, but you got an extra 3 in the denominator in front of sin(30)

- anonymous

You still need to multiply that by 3, and by sin(30)

- anonymous

wait, im confused. so it would be (5)(9.81) / (sin30-.3cos30) but why do we need to multiply it by 3sin30 ?

- anonymous

anyone?

- 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

its all good! and now i get 306 :)

- anonymous

and so for the last one it would be 306(cos)30?

- anonymous

Using 9.81 as g that is correct, and for the second thats almost it.

- anonymous

306 is correct, but this is the work done by the force.

- anonymous

On the last it is asking for the normal, wich is a force, not work.

- anonymous

It is given by Fcos(30) and F is what you were calculating before.

- anonymous

F= that 204 you were calculating.

- anonymous

but Fcos(30) is the normal force?

- anonymous

i got 204 by mistake though

- 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

i tried it again, and still didnt get that :/

- anonymous

im sorry :(

- anonymous

Wait, I said 204 is the force, not the answer, the answer is 204cos(30)

- anonymous

but i dont get 204 at all

- anonymous

The formulas are:\[N=Fcos \theta\]\[F=\frac{mg}{\sin \theta-\mu \cos \theta}\]

- anonymous

The explanation for them is in the begining.

- anonymous

i did that

- anonymous

Where you using that?

- anonymous

yes

- anonymous

Show the numbers again.

- anonymous

5)(9.81) / (sin30-.3cos30)

- anonymous

And you need to multiply all of that by cos(30)

- 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

Yes, but remember what those numbers mean.

- anonymous

i will, im just unsure with how u knew which equations to use

- 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

when i checked the answers, the right answers were 310J... -150J...180 N

- anonymous

310 for a
-150 for b
and 180 for c

- anonymous

i got a similar answer for the first 2 parts, but not for the third!

- 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

why sqrt of 3/2 ?

- anonymous

i think you just mean cos30 because those two are the same things :)

- anonymous

What?? wait, thats what i meant:\[\cos 30°=\frac{\sqrt{3}}{2}\]

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