How Do You Calculate Braking Force Without Kinetic Energy?

  • Thread starter Thread starter rokr32
  • Start date Start date
  • Tags Tags
    Car
Click For Summary
SUMMARY

The discussion focuses on calculating the average braking force exerted on a car that comes to rest over a distance of 15 meters, with an average resistive force of 2.4 x 103N. The correct approach involves using the equation F = ma, while avoiding kinetic energy calculations. The final calculated braking force is confirmed to be 2.6 x 103N, derived from the relationship between resistive forces and the net force acting on the car.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Basic knowledge of forces and free body diagrams
  • Familiarity with resistive forces in motion
  • Concept of average force calculations
NEXT STEPS
  • Study free body diagram techniques for visualizing forces
  • Learn about the implications of signs in vector quantities
  • Explore advanced topics in dynamics, including deceleration calculations
  • Investigate the relationship between distance, force, and acceleration in physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and force calculations, as well as educators looking for examples of problem-solving in dynamics.

rokr32
Messages
5
Reaction score
0

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
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
Replies
12
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K