A 1500 kg automobile is traveling up to 20 degree incline at a speed of 6 m/s. If the driver wishes to stop his car in a distance of 5m,determine the frictional force at pavement which must be supplied by rear wheels.

KE=PE+W(fr)

m•v²/2 = m•g•h +F•s=
=m•g•s•sinα +F•s.
F=m(v²/2s - g•sinα)

To determine the frictional force exerted by the rear wheels of the car, we need to analyze the forces acting on the car.

1. Forces acting on the car:
- Weight: The weight of the car acts vertically downward and can be determined by multiplying the mass of the car (m) with the acceleration due to gravity (g) which is approximately 9.8 m/s^2. W = m * g = 1500 kg * 9.8 m/s^2 = 14700 N.
- Normal force: The normal force is the force exerted by the ground perpendicular to the incline. It can be calculated by taking the component of weight acting perpendicular to the incline. N = W * cos(θ) = 14700 N * cos(20°) = 13887 N.

2. Calculate the component of weight parallel to the incline:
- The component of weight parallel to the incline is given by W_parallel = W * sin(θ) = 14700 N * sin(20°) = 5018.7 N.

3. Calculate the net force acting on the car:
- The net force acting on the car is given by the difference between the parallel component of weight and the frictional force. Net Force = W_parallel - Frictional force.

4. Calculate the deceleration:
- The deceleration of the car can be calculated using the kinematic equation: v^2 = u^2 - 2as, where
- v = final velocity (0 m/s, as the car wants to stop)
- u = initial velocity (6 m/s)
- a = acceleration (deceleration)
- s = displacement (5 m in this case)

Rearranging the equation, we get:-