Calculating Forces: Car Stopping & Box Moving

[claralax]

Forces! please helpp!

1) 2500kg car stops from 14 m/s to full stop in a distance of 25.0 m. what is the coefficient of kinetic friction?q

2ad= v squared - v(starting) squared
2a25 = - (14) squared
50a = -196
a = - 3.92 m/s

i got this far and then to find Ffriction = u(coefficient)x normal force...and here i got lost.

i know that to stop the car a force of - 9800 N is needed. (F = ma = 2500(-3.92) = -9800N

but how do i get the friction of force? can somebody help me...please?

do i use Fg at all? Fg=2500(9.8)=24500N?

and then 2) box of 5 kg is moving at velocity of 3 m/s due to Force of 10N. What is the normal force?

I don't understand normal force. i know its a weight of an object and any other force added to it.

so in this case would it be 10N + 5(9.8)N which equals to 59N?

 

[Yapper]

Is there anymore to number 1? It seems they only give enough information to find the stoping force... Do they give the braking force? if they did then just subtract from the net force which is the 9800N

 

[claralax]

no they don't give any more information.
is number two right though?
or should i use pythagorian theorem for that since the speed is constant? so the net force is equal to zero?

 

[vincentchan]

f_{k} = \mu N , f_{k} is the friction. and what deccelarate the car is friction... N is normal, which is the force the car applied on the floor... after you have these two, finding \mu shouldn't be a problem

 

[Yapper]

He can get normal, but how does he get friction force?with given info Normal force in this case is = to mg

 

[claralax]

could the force the car needs to stop equal the force of friction since it drives at icy road?

 

[vincentchan]

for your question, yes...in your daily life...NO

 

[Yapper]

lol, next time copy whole question.

 

[claralax]

well i wasnt sure...but okay

 

[claralax]

ima real sorry guys...but didnt seem important at the beginning...ay ay

HOW BOUT THE SECOND QUESTION??

 

[Yapper]

Normal force is the force that the surface exerts on the box... if the box is on a flat surface, and the force is being applied perfectly horrizantally than it should just be equal to the force that the box is exerting on the floor caused by gravity, in which case it would be mg

 

[claralax]

thanx a lot...man why can't they explain it clearly like this in the book? why can't my physics teacher teach for a change?...and stop making us study by ourselves from the book...*sigh*...THANX YOU ALL who helped me today.

 

[Zaimeen]

He can get normal, but how does he get friction force?with given info Normal force in this case is = to mg

as vincentchan suggested, friction = \mu R, you can get the frictional force by using Friction = m a where a is the magnitude of the deceleration. There are no other horizontal forces being applied, so, you can easily evaluate for \mu