Question

A 2.1 ? 10³-kg car starts from rest at the top of a 4.1-m-long driveway that is inclined at 16

Answer 1

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