If you're not too busy

[dash00]

Please take a minute to help me out with highschool phys prob.

 

 

A car of mass 1200kg is travelling down a hill with an incline of 1 in 10.

 

Speed = 30 m/s when a constant braking force was applied. The car stopped after travelling 90 m down the incline.

 

What was the magnitude of the constant braking force applied?

 

my answer for it is incorrect . i got 6000N, but i didnt take into account the fact that it is on an incline at all lol, i had a go but wasnt sure how to

 

thanks for your help in advance im sure this will be childs play for most of you!

 

edit: sorry people, i didnt realise there was the 'homework help' section here, i just registered tonight, sorry again.

[Klaynos]

well you'll need to work out the angle between the ground and the incline, using trig.

 

Then if we consider that gravity is pullin down on the car with a force = Mg, and part of this force is in the direction of travel, we can use the angle above to work out how much of this force is in the direction of motion, you'll find if you draw it out that the angle between the Reactant force on the car, and the direction of the force due to gravity is the same as the one between the incline and the ground, so that can tell us that the part of the force in the direction of motion of the car is MgSin(Angle), so the breaking force will also have to compensate for this.

 

 

Sorry I'm not being too clear, don't want to do the whole question for you, and I'm quite tired

[dash00]

Thanks for the reply. i realised thats wat i needed to and worked out the component of the force thats in the direction of travel to be 1200 N (is that right?) but once i'd done that i wasnt sure how to put it all together, so i came here.

 

Could you please tell me how to go about working it out after this stage? i have spent a fair bit of time trying to figure out wat im missing so help would realy be appreciated if you would be so kind.

 

answer provided is 6.1 *10^4 N btw.

 

thanks

[Klaynos]

Well

 

F=sum of all the forces = ma + mgSinTheta

 

ma = 6000 as you got

 

Theta = the angle between the incline and the ground (arctan(1/10))

 

mg = force due to gravity.

 

I got an answer of 7171N which is wrong according to your given answer so I'd probably assume I'm getting something wrong or missing something.

[Primarygun]

I got the same answer

[dash00]

firstly, thanks alot guys for your responses and taking the time and effort to help me out.

 

i got the answer 7200 as well (g= 10 , lazy ) but when i saw how far out i was from the teachers answer i thought it was something more complicated than summing them. Suppose this car was dropped of a cliffe, wouldnt it only require a force of mg = 12000N to stop it? i guess the answer is wrong (61000N)! ( it is just the teacher who writes up the answers on back of sheet, mistakes arent too out of the ordinary)

 

thankyou very much

 

edit: btw, to work out the 6000N i used the motion equation x= (u + v)t

where time was the unknown before using Favg= change in momentum/time

 

did you guys use the same method or a different way?

 

ps. what is a lepton?

[Primarygun]

g= 10

In my region, our syllabus is g=10ms^2.

edit: btw, to work out the 6000N i used the motion equation x= (u + v)t

where time was the unknown before using Favg= change in momentum/time

I use v^2=u^2+2as to find out the a first.

[dash00]

being a smart ass accomplishes what exactly?

[Klaynos]

In my region' date=' our syllabus is g=10ms^2.

 

I use v^2=u^2+2as to find out the a first.[/quote']

 

We use anything from 10 to actually working out what it should be due to altetude etc...

 

I used the v^2=... as well

 

Then the joys of F=ma