How Do You Calculate Braking Force Without Kinetic Energy?

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To calculate the average braking force without using kinetic energy, the problem states that a car comes to rest in 15 meters, with an average resistive force of 2400N. The initial approach involved setting up an equation based on forces, but there was confusion regarding the use of the term "15F." The discussion emphasizes the importance of considering the direction of forces, as signs indicate their orientation in a coordinate system. Ultimately, the correct average braking force is confirmed to be 2.6 x 10^3N. A suggestion is made to draw a free body diagram to visualize the forces acting on the car.
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Homework Statement


1o27w6.jpg


The driver applies the brakes and the car comes to rest in 15m. Calculate the average braking force exerted on the car in coming to rest.

The average resistive force acting on the car is 2.4 x 103N

Homework Equations


F=ma
SUVAT maybe?

The Attempt at a Solution


When using KE = 1/2mv2, the problem is easy to solve. However, I'm not allowed to use this. Therefore, I have NO IDEA how to solve this.

One approach I took is, given that the average resistive force is 2400N, I did:
15F + 2400 = 0
F = -160N.

Then, given that the signs are irrelevant, I can add 160N to 2400N and get 2560N which rounds to 2.6 x 103N.

By the way, the answer should be 2.6 x 103N

Thanks :D
 
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rokr32 said:

Homework Statement


1o27w6.jpg


The driver applies the brakes and the car comes to rest in 15m. Calculate the average braking force exerted on the car in coming to rest.

The average resistive force acting on the car is 2.4 x 103N

Homework Equations


F=ma
SUVAT maybe?

The Attempt at a Solution


When using KE = 1/2mv2, the problem is easy to solve. However, I'm not allowed to use this. Therefore, I have NO IDEA how to solve this.

One approach I took is, given that the average resistive force is 2400N, I did:
15F + 2400 = 0
F = -160N.
Um, where does the 15F come from?
Then, given that the signs are irrelevant, I can add 160N to 2400N and get 2560N which rounds to 2.6 x 103N.
Signs are never irrelevant! They tell you which direction (with respect to your chosen coordinate system) things are acting or moving.
By the way, the answer should be 2.6 x 103N

Thanks :D

Can you list all the forces acting on the car and their directions? Draw a free body diagram.
 
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