Is there anything missing in this motion equation?

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The discussion revolves around calculating the average braking force of a car using motion equations. The initial query presents a scenario where a 1100-kg car decelerates from 27 m/s to a stop over 578.0 m, seeking to confirm the average braking force of -690 N. Participants suggest using Newton's second law and the equation of motion, ultimately leading to the conclusion that the average braking force can also be derived through energy conservation principles. The original poster successfully arrives at the correct answer by applying the equations directly without needing the final derived formula. The conversation concludes with appreciation for the assistance received.
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hello
have found this interesting question (interesting to me)... along with an equation that is supposed to allow the answer... but I plug in the data and never get the same result as is correct.

could someone set it out with the data so it flows to give -690N (if correct)... thanks
(not an exam question or anything... I found it and would like it as a model to find average brake force... cheers)



A 1100-kg car traveling at 27 m/s starts to decelerate and comes to a complete stop in 578.0 m. What is the average braking force acting on the car?
answer: (a) -690 N (b) -340 N (c) -410 N (d) -550 N

the answer is said to be (a) -690N
and the suggested equations are these:

Use Newton's second law in the form:
F = ma

And this equation of motion:
v2 = u2 + 2as

Where:
m = mass
F = braking force
v = final velocity
u = initial velocity
s = displacement

You end up with an equation like this:
F = (mv2)/2s

and the answer is -690 N.

.......
appreciate thoughts, thanks
 
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F = (mv^2)/2s gives 693 N.You could have gotten the same answer by using conservation of energy:

Loss of KE by car = work done by friction force

1/2(mv^2)=Fs
 
rock.freak667 said:
F = (mv^2)/2s gives 693 N.


You could have gotten the same answer by using conservation of energy:

Loss of KE by car = work done by friction force

1/2(mv^2)=Fs

Thanks... gotit in end...
didn't bother with the end equation... simply used first 2 equations in own right:

* started with v2=u2+2as
* then plugged that into F=ma
* this got me to the answer

All done
Thanks for input, much appreciated
 
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