- #1
PascalPanther
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I am suppose to find the initial velocity of a car before it skids to a stop.
The car begins it's stop right as it is going up a road that is 20 degrees above the horizontal. The car makes a 15.2 m skid mark before it stops. The coefficient of kinetic friction is 0.60, and static friction is 0.80. The mass of the car and its driver is 1630kg.
Now this is my assumption, that since the car left a skid mark, that would mean the wheels aren't turning. Which I don't think matters, since it would still be kinetic even if the wheels were turning (?) Since I am stopping, there is no other positive force working against the force of friction. My final velocity is 0 since I stop.
First thing I should do is find the force of friction.
Force of friction = (coef) (normal force)
The normal force is the force parallel to the plane, so it is m*g*cos20
F(k) = (0.60) (1630kg*9.8m/s^2) = 9010 N
Next:
F = ma; 9010N = 1630kg(a) = 5.53 m/s^2
I think I can use kinematics now, and:
vf^2 - vi^2 = 2a(x)
vi^2 = 2(5.53 m/s^2)(15.2m)
vi = 13.0 m/s
Does that seem about right? Or am I suppose to use the static friction in there somehow?
The car begins it's stop right as it is going up a road that is 20 degrees above the horizontal. The car makes a 15.2 m skid mark before it stops. The coefficient of kinetic friction is 0.60, and static friction is 0.80. The mass of the car and its driver is 1630kg.
Now this is my assumption, that since the car left a skid mark, that would mean the wheels aren't turning. Which I don't think matters, since it would still be kinetic even if the wheels were turning (?) Since I am stopping, there is no other positive force working against the force of friction. My final velocity is 0 since I stop.
First thing I should do is find the force of friction.
Force of friction = (coef) (normal force)
The normal force is the force parallel to the plane, so it is m*g*cos20
F(k) = (0.60) (1630kg*9.8m/s^2) = 9010 N
Next:
F = ma; 9010N = 1630kg(a) = 5.53 m/s^2
I think I can use kinematics now, and:
vf^2 - vi^2 = 2a(x)
vi^2 = 2(5.53 m/s^2)(15.2m)
vi = 13.0 m/s
Does that seem about right? Or am I suppose to use the static friction in there somehow?