Where brick hits ground after sliding off a roof w/ friction

AI Thread Summary
A brick sliding off a roof at a height of 7.0 m, with a friction coefficient of 0.28, is analyzed for its landing point after sliding down to a height of 4.0 m. The discussion involves calculating the acceleration down the roof using the equation ma = mgsin(30) - frictional force, where the frictional force is derived from the coefficient of kinetic friction. Participants clarify that the frictional force is calculated as Ff = 0.28 * mgcos(30). The conversation also shifts to determining the speed of the brick as it leaves the roof. The focus remains on applying physics principles to solve the problem accurately.
isabelrose
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Homework Statement


Brick is sliding off a roof, starts at the peak of the roof (7.0 m above ground), slides down roof with kinetic friction value of 0.28, very edge of roof is 4.0 m above ground. Where does the brick land on the ground?
Please see image - http://imgur.com/wNdtlWD

Homework Equations

The Attempt at a Solution


I attempted the solution by: [/B]
drawing force diagram of brick (@ = angle of incline, 30 degrees) which gives the acceleration down (acceleration down the roof = positive direction) the roof by:
ma = mgsin@ - frictional force
ma = mgsin30 - 0.28

Now should I just continue to solve for the acceleration and then treat this as a projectile? If so, can someone walk me through it?

Thanks!
 
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0.28 is not the frictional force. It is the coefficient of kinetic friction. How do you obtain the frictional force from the coefficient?
 
haruspex said:
0.28 is not the frictional force. It is the coefficient of kinetic friction. How do you obtain the frictional force from the coefficient?
Oh yeah!
The frictional force (Ff) = coefficient of kinetic friciton * Fn = 0.28 * mgcos@
 
isabelrose said:
Oh yeah!
The frictional force (Ff) = coefficient of kinetic friciton * Fn = 0.28 * mgcos@
Good.
So, at what speed does it leave the roof?
 

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