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DawG_M
11-29-2004, 11:41 PM
Does anybody know how to do this problem? I really dont know how to even start.

A block slides down an inclined plane 9m long that makes an angle of 38 degrees with the horizontal. The coefficient of sliding friction is 0.25 If the block starts from rest, find the time required for it to reach the foot of the plane

Robert00000
11-29-2004, 11:43 PM
Does anybody know how to do this problem? I really dont know how to even start.

A block slides down an inclined plane 9m long that makes an angle of 38 degrees with the horizontal. The coefficient of sliding friction is 0.25 If the block starts from rest, find the time required for it to reach the foot of the plane
:huh: .........

DarthInsinuate
11-29-2004, 11:44 PM
i think you need to work out the vertical or horizontal component or both and throw in 9.8 somewhere

TheDave
11-29-2004, 11:44 PM
you missed anything?

DawG_M
11-29-2004, 11:50 PM
thats all the info i got, im thinking that its impossible to do

TheDave
11-29-2004, 11:52 PM
nothing from a previous question?

DawG_M
11-29-2004, 11:56 PM
no this is the only problem the teacher gave us

ziggyjuarez
11-29-2004, 11:56 PM
wow
make me feel dumb.I dont even know 5x6 and thats no joke :<

TheDave
11-30-2004, 12:01 AM
The coefficient of sliding friction is 0.25 If the block starts from rest


i dont get it. if this is supposed to multiply or something it never moves :blink:

*goes to find what coefficiant means*

TheDave
11-30-2004, 12:07 AM
if 0.25 means metres per second it takes 36 seconds. but it cant be, otherwise it wouldn't state starting from rest. maybe if there was more info i could work out the acceleration. :unsure:

TheDave
11-30-2004, 12:24 AM
maybe it means mps+0.25m=mps

manker
11-30-2004, 12:27 AM
Does anybody know how to do this problem? I really dont know how to even start.

A block slides down an inclined plane 9m long that makes an angle of 38 degrees with the horizontal. The coefficient of sliding friction is 0.25 If the block starts from rest, find the time required for it to reach the foot of the planeYou'd need the weight of the block to determine this.

I'm looking at it and logically a housebrick will take longer to travel down the 9m slope than a bit of lego :unsure:

Anyway, I think you'll need to use trigonometry too so look at the question in context with what else you're learning in school. I suspect the answers will be there if you look.

TheDave
11-30-2004, 12:30 AM
what about greased brick vs sticky lego :smilie4:

manker
11-30-2004, 12:32 AM
Greased brick :01:

ZaZu
11-30-2004, 12:46 AM
Guy hit me inna head with a greased brick once....once

DawG_M
11-30-2004, 12:59 AM
There must be some way to do this w/o mass since he didnt give it to us.

DarthInsinuate
11-30-2004, 01:31 AM
i don't think you need the mass because you can work out time from just acceleration and starting speed

edit: :dry: and distance

TheDave
11-30-2004, 01:34 AM
i dont think we have acceleration

DarthInsinuate
11-30-2004, 01:37 AM
acceleration due to gravity, argh i hate this thread, i did slopes and friction for my coursework i should know this, but i don't

Barbarossa
11-30-2004, 11:23 AM
Heres some equations that may jog your memory:

distance = 1/2 * acceleration * time squared.

Therefore, time = square root of twice the distance divided by the acceleration. [t = SQRT(2d/a)]

So we need to find the acceleration, but all we've been given is the friction coefficient and the angle of incline. We haven't been given mass, so I must assume that it somehow cancels out in a minute.. (?)

Anyway, "Normal" Force = Mass * Acceleration Due to Gravity * Cosine(Angle of Incline)

Force of Friction = "Normal" Force * Friction Coefficient

Now the force of friction acts in the opposite way to the "Normal" force, so the net force = Normal Force - Force of Friction, which basically is the Normal Force times (1 minus the friction coefficient) ..

Therefore, the acceleration down the slope is Net Force / (Mass * Sine(Angle of Incline))

If we combine all this we get: A = M*G*Cos(S)(1-U)/M*sin(S), and so yes, M cancels out, so A=G*Cos(S)(1-U)/Sin(S)

G=Gravity=9.8
Cos(S)=Cos(38) = 0.7880107536
Sin(S)=Sin(38) = 0.6156614753
U=Friction Coefficient=0.25

Therefore A=Acceleration down the slope = 9.407570997

Therefore, T= SQRT(2d/A)

A=Acceleration=9.407570997
D=Distance=9

T=1.383239876 SECONDS.

Fuck me, that was quite hard, it's over a decade since last did an exam..

Don't come crying to me if it's wrong. I'd put money on it..
:ph34r:

Afronaut
11-30-2004, 12:17 PM
Remember to ad a link to FST on the Exam,
or better yet, straight to this thread.

:lol: