# Thread: Physics - friction up a slope

1. ## 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

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?

2. ## Resolving forces

Hello Intrusion
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

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

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

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

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:

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

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

Can you complete it now?