How Much Additional Braking Force is Needed to Stop the Car at Point B?

[soul5]

Homework Statement

A passenger car has a speed of 16.8 m/s when it reaches point A. The average frictional force between the car and the track from point A to B is 250N. At point B in addition to this frictional force, the car's breaks are applied so that the passenger car stops at point B.

Question: What additional force that must be applied by the breaks between point A and B in order to stop the car at point B?

Homework Equations

Ek=1/2mv^2

The Attempt at a Solution

EK is all I know.

 

[AvroArrow]

Would I use EK at point A, subtract the Ek at point B and then solve for the force? EK = 1/2mv^2 F=m(v2-v1)/t

 

[Moetasim]

I would like to clarify that the law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In this scenario, the initial kinetic energy of the car at point A is converted into other forms of energy, such as heat and sound, due to the frictional force acting on the car.

To calculate the additional force required to stop the car at point B, we can use the equation for work, W = Fd, where W is work, F is force, and d is distance. The work done by the brakes must be equal to the initial kinetic energy of the car, which is given by Ek = 1/2mv^2. Therefore, the additional force required can be calculated as F = (1/2mv^2)/d.

However, it is important to note that the frictional force also contributes to slowing down the car, so the total force applied by the brakes will be the sum of the additional force calculated and the frictional force of 250N. This additional force will vary depending on the distance between points A and B, but it is necessary to ensure the car comes to a complete stop at point B.