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

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

how do we find the change in x?

so instead of N should i plug in Fcosθ

and what equation would we use for that?

a little, not really because i havent used it before

or at least i dont believe i have

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

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

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

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

oh okay!

Any doubts?

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

Oh, it would be that simple?

Yes, that simple

ok... ur good learner .,

haha thank you!

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

Yes!

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

I got approximately 312 using g=10.

the mass would be 5 correct?

yes

idk what i did wrong :/

Are you talking about part (a)?

yes

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

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

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

anyone?

its all good! and now i get 306 :)

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

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

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

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

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

F= that 204 you were calculating.

but Fcos(30) is the normal force?

i got 204 by mistake though

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

im sorry :(

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

but i dont get 204 at all

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

The explanation for them is in the begining.

i did that

Where you using that?

yes

Show the numbers again.

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

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

Yes, but remember what those numbers mean.

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

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

310 for a
-150 for b
and 180 for c

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

why sqrt of 3/2 ?

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

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