maybe it means mps+0.25m=mps
You'd need the weight of the block to determine this.Originally Posted by DawG_M
I'm looking at it and logically a housebrick will take longer to travel down the 9m slope than a bit of lego
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.
what about greased brick vs sticky lego
Guy hit me inna head with a greased brick once....once
If you attack the establishment long enough and hard enough, they will make you a member of it.
-- Art Buchwald --
There must be some way to do this w/o mass since he didnt give it to us.
i don't think you need the mass because you can work out time from just acceleration and starting speed
edit: and distance
Last edited by DarthInsinuate; 11-30-2004 at 01:32 AM.
The Sexay Half Of ABBA And Max: Freelance Plants
i dont think we have acceleration
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
The Sexay Half Of ABBA And Max: Freelance Plants
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..
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