How Do Brakes Apply Stopping Force in Physics Problems?

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Brakes apply stopping force by converting kinetic energy into work, as described by the work-energy theorem. In the provided example, a boy on a bicycle comes to a stop, and the change in kinetic energy is calculated to be -7312.5 J. This negative value indicates energy loss due to braking. The work done by the brakes equals this change in energy, allowing for the calculation of the average force applied. Understanding these concepts is crucial for solving physics problems related to stopping force.
guhroovy_dainzaar
I get most of the things in my textbook, such as mgh,Fs and 1/2mv squared. But I don't understand things about stopping force and stuff.

heres a practice question i got(i have a test 2morrow):

A boy of mass 50 kg rides on a bicycle of mass 15 kg at a speed of 15 m/s. He suddenly applies the braked and brings the bicycle to a stop in 25m. What force did the brakes apply to the wheels as he was stopping?

I really don't understand how to do this... please help!
 
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Well, what was the change in energy of the bicycle? The only energy that is changing here is its kinetic energy, so you can compute the change in energy. (It's negative, btw... always be careful to keep your signs straight!)

The work-energy theorem says that the change in energy of a system is equal to the work applied to the system.

So now that you know the work done by the brakes (and the distance through which it was applied), you can compute the average force applied by the brakes!
 
?

i is still confused... how do i actually calculate it?
 
Can you calculate the initial and final kinetic energy?
 
sorta ish

I can calculate that the enitial kinetic energy is 7312.5 but after that i don't know wot 2 do
 
Well, the final kinetic energy is 0, right? (speed is 0)


The change in energy, ΔU, is equal to the final energy minus the initial energy. Since kinetic energy is all that is changing, we know that:

ΔU = 0 J - 7312.5 J= -7312.5 J


The work-energy theorem tells us that the work done on the system must therefore be -7312.5 J... can you take it from there?
 
thanx

thank u so much... I mite actually pass my test tomorrow!
YAY:smile: :wink:
 
Good luck!
 
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