# Kinematics - Find static friction coefficient of a coffee cup

1. Jan 16, 2008

### Elvis

Hi !

Can anyone help me I need help ......

Problem :
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1- A coffee cup on the dashboard of a car slides forward on the dash when the driver decelerates from 40 km/h to rest in 3.5 s or less, but not if he decelerates in a longer time. What is the coefficient of static friction between the cup and the dash ?
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2- An 18-kg box is released on a 37*(grade) incline and accelerates down the incline at 0.270 m/s(square). Find the friction force impeding its motion. How large is the coefficient of friction
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Thanks !!!!

Last edited: Jan 16, 2008
2. Jan 16, 2008

### WiFO215

what have you tried?

3. Jan 16, 2008

### Elvis

Problem 1:
I now that :

F = F
uN = ma
umg = ma
ug = a
u(9.8) = (40km/h * 1h/3600s * 1000m/km)/3.5s
u = 0.324

Is this correct ?

Can anyone help me for the problem 2 ?

Thanks

Last edited: Jan 16, 2008
4. Jan 16, 2008

### Elvis

Problem 2
I have solved these problem ::::::

The weight of the box is (18 kg)(9.80 N/kg) = 176.4 N
The component of the weight parallel to the plane is
(176.4 N)sin(37.0o) = 106.16 N

The component of the weight perpendicular to the plane is
(176.4 N)cos(37.0o) = 140.88 N
So now our expression of Newton's second law has the parallel component of gravity of 106.16 N down (-) the plane, and assuming it is moving down the plane as well as accelerating down the plane, the force of friction F is acting up (+) the plane, and the acceleration is .270 m/s/s down (-) the plane, so:
<F - 106.16 N> = (18.0 kg)(-.270 m/s/s)
F = 101.3 N = 101 N, and the coefficient of friction is given by
Ffr = mkFn
The Normal force in this case is the perpendicular component of gravity 140.88 N, so we have:
Ffr = mkFn
101.3 N = mk(140.88 N)
So mk = .719

5. Jan 16, 2008

### WiFO215

There you go!You're done!