A 2172.5 kg car is coasting along a level road at 34.7 m/s.
A constant braking force is applied,such that the car is stopped in a distance of 50.5 m.
What is the magnitude of the braking force? Answer in N.
Here is my way, but im not sure if im correct:
Vf^2=Vi^2+2a(delta position)
0=34.7^2+2a(50.5)
=11.92 m/s^2
2172.5(11.92)=25,896.2 N
I used the same Eq and procedure you
used and I got 25899.9N. So, i agree with your answer.
thanks!!
Well, your calculations seem to be on the right track. However, let me add a touch of humor to your answer:
To find the magnitude of the braking force, we'll do some math that even the car itself would admire. According to your calculations, the car experiences a deceleration of 11.92 m/s^2. This means that the car must really appreciate the laws of physics, as it willingly slows down – unlike some drivers on the road!
Now, multiplying the mass of the car (2172.5 kg) by the deceleration (11.92 m/s^2), we find that the magnitude of the braking force is a whopping 25,896.2 N. That's a force even strong enough to make a clown jump out of their tiny car!
So, to summarize, the magnitude of the braking force to stop the car in a distance of 50.5 m is approximately 25,896.2 N. Keep up the good work, mathematician (and comedian)!
Your calculation is almost correct! However, there is a small mistake in your calculation of the acceleration.
The correct equation to use is:
Vf^2 = Vi^2 + 2a(delta position)
0 = (34.7 m/s)^2 + 2a(50.5 m)
Simplifying the equation gives us:
0 = 1204.09 m^2/s^2 + 101a
101a = -1204.09 m^2/s^2
a = -11.92 m/s^2
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity.
Now, to calculate the magnitude of the braking force, we can use Newton's second law:
F = ma
F = (2172.5 kg)(-11.92 m/s^2)
F ≈ -25,896.2 N
Therefore, the magnitude of the braking force is approximately 25,896.2 N.