Is there anything missing in this motion equation?

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SUMMARY

The discussion centers on calculating the average braking force of a car using Newton's second law and the equations of motion. A 1100-kg car traveling at 27 m/s decelerates to a stop over a distance of 578.0 m, resulting in an average braking force of -690 N. The participants confirm the validity of the calculation using both the equation F = (mv²)/2s and the conservation of energy principle, demonstrating that the correct answer is indeed -690 N.

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  • Understanding of Newton's second law (F = ma)
  • Familiarity with kinematic equations, specifically v² = u² + 2as
  • Knowledge of energy conservation principles in physics
  • Basic algebra for manipulating equations
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stephenn
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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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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