# Physics - friction up a slope

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• February 18th 2009, 06:30 PM
Intrusion
Physics - friction up a slope
Problem:
Lindsey's car is driving up a hill when her book bag, which she forgot on the roof, falls off. The bag has a mass of 19kg and slides up the hill 60 meters from its initial speed of 20m/s to rest. If the hill climbs at an angle of 30 degrees then what was the coefficient of friction?

My attempt:

Fn = 19*9.8cos30 = 161N

Vf^2 = Vi^2 + 2ad
0 = 202 + 2a(60)
-400 = 120a
a = -3.33 m/s2

F = ma
F = 19 * -3.33 = 63.27N

Ff = μmgcosx
63.27 = μ19*9.8*cos30
μ= .393

Is this correct? Is the parallel component required in this case? In what cases is it required?
• February 18th 2009, 10:44 PM
Grandad
Resolving forces
Hello Intrusion
Quote:

Originally Posted by Intrusion
Problem:
Lindsey's car is driving up a hill when her book bag, which she forgot on the roof, falls off. The bag has a mass of 19kg and slides up the hill 60 meters from its initial speed of 20m/s to rest. If the hill climbs at an angle of 30 degrees then what was the coefficient of friction?

My attempt:

Fn = 19*9.8cos30 = 161N

Vf^2 = Vi^2 + 2ad
0 = 202 + 2a(60)
-400 = 120a
a = -3.33 m/s2

F = ma
F = 19 * -3.33 = 63.27N

Quote:

Originally Posted by Intrusion
Ff = μmgcosx
63.27 = μ19*9.8*cos30

μ= .393

Is this correct? Is the parallel component required in this case? In what cases is it required?

Thanks for showing us your working.

The diagram shows the forces acting on the book.

Resolve at right angles to the plane:

$N - mg\cos 30^o = 0$

So your first equation is correct.

And you have correctly calculated the acceleration (although you have rounded off a bit too much; the answer is more accurately 63.33.)

But when you resolve down the plane:

$F + mg\sin 30^o = ma$

So your 'friction force' equation is wrong. You need to take the component of the weight into account.

Can you complete it now?

Grandad