force

[princesspriya]

1. The problem statement, all variables and given/known data
use to answer 1 and 2. A truck driver slams on the brakes and skids to a stop through a displacement of x.
1.) if the truck's mass doubles, find the truck's skidding distance in terms of x.(hint:increasing the mass increases the normal force.)
a. x/4
b. x
c. 2x
d. 4x

2.) If the truck's initial velocity were halved, what would be the truck's skidding distance?
a. x/4
b. x
c. 2x
d. 4x

wudnt both of them be x/4 because if u increase the mass than the acceleration would decrease so the distance would decrease also and the only one less than x is x/4 and the same thing with initial velocity. if the ini. vel. is less than the acceleration would be less. but my teacher said that my answers were wrong. so please help.


2. Relevant equations



3. The attempt at a solution

[Snazzy]

For the first question, -Ff = ma. If Ff is proportional to the normal force, which is proportional to the mass in this case, does increasing the mass change the acceleration?

I agree with your answer for the 2nd question, but your answer for 2 is for the wrong reasons.

[princesspriya]

it should coz more mass would mean less acceleration by newtons second law and y would my reasoning be wrong for 2nd question? also doesnt Fnet=ma not Ff?

[Snazzy]

But F_f=\mu mg and -F_f=ma

[princesspriya]

doesnt Ff= umgcosX becuase Fn is mgcosX and than you have to find all the netforces to say is equal to ma

[Snazzy]

The normal force is equal but in an opposite direction to the gravitational force in this question. The only force acting in the direction of acceleration is the frictional force when a truck slams its breaks.

[princesspriya]

right?? so that means that more mass less acceleration

[Snazzy]

If -F_f=-\mu mg=ma what happens if you increase the mass? How does it affect the acceleration?

[princesspriya]

the acceleration decreases.

[Snazzy]

Why would it decrease if the mass cancels out?

[princesspriya]

because lets say that the Ff was like 40N then changing the values of mass would decrease the acceleration.

[tmoan]

if i may the acceleration you are talking about is the one that is stopping the truck. therefore logically and without all the equations the truck takes more time to stop and therefore more displacement.
your logic is right and u understood the question. friction always acts opposite to motion and so it stops the truck.
this acceleration is negative w.r.t. direction of motion....
now if you apply your equations it should give 4x

[princesspriya]

tmoan are you talking about the 1st or the 2nd problem because i thought it would be x/4 for both of them.

[tmoan]

the first where the mass doubles

[kamerling]

if i may the acceleration you are talking about is the one that is stopping the truck. therefore logically and without all the equations the truck takes more time to stop and therefore more displacement.
your logic is right and u understood the question. friction always acts opposite to motion and so it stops the truck.
this acceleration is negative w.r.t. direction of motion....
now if you apply your equations it should give 4x

But the friction depends on the normal force, wich depends on the mass of the truck.

A truck with 2x the mass will have 2x the normal force and thus 2x the friction force.
the acceleration will stay the same.

[tmoan]

listen if there was no friction it slams intu a wall right in a millisecond. say the truck was moving at 20 km/h. in a millisec that became 0 km/ h. so it has a very large acceleration directed opposite of motion.
on the other hand take the friction case. friction doesn't stay the same because the sum of friction force during the whole stopping period will be constant from v =20 to v = 0.
but the difference is in acceleration.
you yourself said the acc decreases and you are right.

[tmoan]

ur right the acceleration stay the same.
but since the friction force doubles.
ur right

[tmoan]

but try to think of it this way if the truck was at a slow speed and you stood infront of it trying to stop it while walkin backwards with it. and u can only exert a constant force on the truck (ur the friction)
the truck will exert a force on ur hand.
if the truck had 2x mass it will exert more force on ur hand so you will take more time to stop such a force right=> more distance.
sorry i can't explain by simple equations.

[princesspriya]

lolzz i m soo lostt.. because more mass means less acceleration which means less displacement

[tmoan]

hehe

[princesspriya]

lolzz ... lmao greatt ahhHH!!!!! i have no cluee wat i m doingg!!! i need a teacher who teaches instead of tellin us stories about his life and cracking lame jokes and that hehe was not cute!! lolz

[tmoan]

you just proved me wrong and disproved yourself again.
m1 and m2
friction force
f1 = k N1 ;N1 = - m1g
f2 = k N2 ;N2 = - m2g = -2m1g
-kg is the acceleration of the friction at all times
ok

[tmoan]

ya i get that in college.Alot

[princesspriya]

right.. do u teach?

[tmoan]

nope i am a student 2nd year physics but my doctors are dumb. and those are PhD holders ...
bunch of crap

[tmoan]

i was actually preparing a small project on maxwell so i registered here. and stumbled on u by accident lol

[Snazzy]

Okay, if you increase the mass, you also increase the normal force so that the frictional force increases as you increase the mass, and there is no change in the acceleration regardless of the mass.

-\mu mg=ma
-ug = a

The acceleration is independent of the mass.

[princesspriya]

lmao i m not like lil puppet dat u can stumble on mee !!!!! lolzz ANYWAYSS THANX ALOT EVERYONE FOR CONFUSING ME!!! lmao now i feel stupid

[princesspriya]

Okay, if you increase the mass, you also increase the normal force so that the frictional force increases as you increase the mass, and there is no change in the acceleration regardless of the mass.

-\mu mg=ma
-ug = a

The acceleration is independent of the mass.

thats not what newton's second law states and where did u get those equations from.

[tmoan]

hehe i didn mean it that way really. i just tried to help i pressed on ur force tab and there u were wit the question
anyways i was just tryin to help... take care

[princesspriya]

now helpp!!! nooo lolzz dnt go lmao since you have the right answer i think.

[Snazzy]

The frictional force is dependent on the mass, the coefficient of static friction, and the acceleration due to gravity, but the deceleration of the truck isn't dependent on the mass AS YOU CAN SEE WITH THE EQUATION. If you can't see that through the cancelling of the mass on both sides of the equation, then you can't move on and your answer will be wrong just as it is right now, with your teacher telling you that it is wrong.

[princesspriya]

but would nt the equations be F=ma and F=2ma and that cancels out the ma leaving behind a 2?

[tmoan]

ya so the force doubles

[Snazzy]

But what is F? F is the frictional force. There are no other forces acting in the direction of motion except the frictional force as I've stated numerous times before. What is the frictional force? It is uF_n where u is the coefficient of friction. What is F_n? It is equal to mg.

Put it all together and you get -umg = ma.

[tmoan]

no no ur wrong f1 = ma
f2 = 2ma
f1 is not equal to f2

[Snazzy]

Okay, I give up. If anyone wants to take over, go ahead.

[princesspriya]

ok i plugged in numbers and i got that the displacement would be 2x

[tmoan]

ok then 2x when i said 4 x i was still thinkin acc would change

[princesspriya]

so its 2x?? YAYYY!! i understood something finallY!!! awww snazzyy gives up on me :(

[tmoan]

and acc's time squares so..
so u got it now

[Snazzy]

NO, IT IS NOT 2X. THE MASS CANCELS OUT ON BOTH SIDES OF THE EQUATION.

THIS EQUATION, RIGHT HERE:

-\mu mg = ma

-\mu g =a

THE DECELERATION DOES NOT DEPEND ON THE MASS.

Does that help?

[princesspriya]

hmm... not really nd omg relaxxx ur typing in caps calmmm downnnnnn

[Snazzy]

Well then, you were destined to get this question wrong.

[tmoan]

wait a sec let me solve it a bit

[princesspriya]

http://img171.imageshack.us/my.php?image=sfsdfdffe7.jpg
thats the work