My own question I thought of this morning. Not sure if I'm right.

[pfaction]

Starting from rest, assuming no friction, how long would it a mass of M at the top of the incline take to get to the bottom of the inclined plane shown above?

So I'm not sure exactly how to approach it here. On one hand, I know it's not just 1/2gtsq=h, gravity is mgsinO and is towards the angle. I thought about using d=1/2xt^sq but I don't think that's right either. So then in a convoluted way; I did:
vfsq = 2(mgsinO)(d)
vf=vo+at -> rad[2(mgsinO)(d)] / mgsinO = t

Am I right, wrong, easier way, etc.

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[MedPR]

Starting from rest, assuming no friction, how long would it a mass of M at the top of the incline take to get to the bottom of the inclined plane shown above?

So I'm not sure exactly how to approach it here. On one hand, I know it's not just 1/2gtsq=h, gravity is mgsinO and is towards the angle. I thought about using d=1/2xt^sq but I don't think that's right either. So then in a convoluted way; I did:
vfsq = 2(mgsinO)(d)
vf=vo+at -> rad[2(mgsinO)(d)] / mgsinO = t

Am I right, wrong, easier way, etc.

Depends on what they give you.

 

[milski]

There are two forces acting on the potato - weight and normal from the ramp. The component of weight that cancels with the normal from the ramp is mg.cosθ, the component parallel to the ramp is mg.sinθ. The acceleration is then g.sinθ and the whole distance is d or h/sin θ.

d=g/2.sinθ/2*t^2=h/sinθ
t^2=h/(sinθ^2 * g/2)
t=sqrt(2h/g)/sinθ

 

[pfaction]

Well this is a free standing conceptual at its finest IMO. I'll read over your answer milski and hopefully it's the same as mine.

 

[milski]

Well this is a free standing conceptual at its finest IMO. I'll read over your answer milski and hopefully it's the same as mine.

I'd comment on yours but I don't get exactly what you mean by "rad[2(mgsinO)(d)] / mgsinO"

sinO is sinθ, but what is rad[]?

 

[pfaction]

Radical the entire thing. I'll write it out in more clear terms.

vf^2 = v0^2 + 2(a)(d)
vf = sqrt (2)(mgsinO)(d)

vf = v0 + at
sqrt[2(mgsinO)(d)] / (mgsinO) = t

 

[milski]

Radical the entire thing. I'll write it out in more clear terms.

vf^2 = v0^2 + 2(a)(d)
vf = sqrt (2)(mgsinO)(d)

vf = v0 + at
sqrt[2(mgsinO)(d)] / (mgsinO) = t

That's the same as sqrt(2d/(mg.sin&#952)=t

It's close but the m needs to go away - the result should not depend on the mass (and the units don't fit anyway).

 

[pfaction]

Yes, I agree, i just noticed I substituted FORCE of gravity rather than pure accereation due to gravity gsinO

All in all I'm so glad I got it right!!

 

[Danlee07]

Would help if this forum supported LATEX...

 

[milski]

Would help if this forum supported LATEX...

I have been bitching about not being able to type formulas for a long time. At least I figured out how to type greek letter. Not much but still better than nothing.