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(fr) •s,
F(fr) =m(v²/2•s – g/ sinα)

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

Please help me to elaborate the answer as i tried but couldn't

To determine the frictional force needed to stop the car, we can use the principle of work-energy. The work done by the frictional force will be equal to the change in the car's kinetic energy.

The kinetic energy of the car is given by the formula: KE = 1/2 * m * v^2

Where:
m = mass of the car = 1500 kg (given)
v = velocity of the car = 6 m/s (given)

Substituting the values into the formula, we have:
KE = 1/2 * 1500 kg * (6 m/s)^2
KE = 1/2 * 1500 kg * 36 m^2/s^2
KE = 27000 kg m^2/s^2

Now, let's calculate the work done by the frictional force. The work done by a force can be calculated using the formula: W = F * d * cos(θ)

Where:
W = work done by the force (unknown)
F = magnitude of the force (unknown)
d = distance over which the force acts = 5 m (given)
θ = angle between the force and the direction of motion = 0 degrees (since the force is opposing the motion, the angle is 0)

Substituting the values into the formula, we have:
27000 kg m^2/s^2 = F * 5 m * cos(0 degrees)
27000 kg m^2/s^2 = 5F

Dividing both sides of the equation by 5, we get:
5400 kg m^2/s^2 = F

Therefore, the frictional force needed to stop the car is 5400 N.