A 2.1 ? 103-kg car starts from rest at the top of a 4.1-m-long driveway that is inclined at 16
The incline decreases the acceleration using this
relationship.
a = g*sin(theta) where theta is the angle with the horizontal (16
degree in this case).
a = 9.8 * sin(16) = 2.7 m/s^2
a = 2.7 m/s^2
m = 2.1 * 10^3 kg
F_downward = m*a = 2.7 * 2.10*10^3 = 5.67 * 10^3 N
The frictional force = 4.0 *10^3 N
Net force = F_downward - Inclined force.
Net Force = 5.67 * 10^3 - 4 * 10^3 = 1.67 * 10^3 N
From the net force we can get an acceleration.
a = F / m
F = 1.67 * 10^3 N
m = 2.1 * 10^3 kg
a = 1.67 *10^3 / 2.10 * 10^3 = 0.795 m/s^2
You have to make the assumption that this car started from
zero.
vi = 0
a = 0.795 m/s^2
d = 4.1 m
vf = ???
vf^2 = vi^2 + 2*a * d
vf^2 = 2 * 0.795 * 4.1
vf^2 = 6.519
vf = 2.56 m/s
so the speed of the car at the bottom of the driveway is 2.56 m/s
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