more physics

[Gibson486]

here is question I am having trouble on

Police lieutenants, examining the scene of an accident involving two cars, measure the skid marks of one of the cars, which nearly came to a stop before colliding, to be 20 m long. The coefficient of kinetic friction between rubber and the pavement is about 0.7. Estimate the initial speed of that car.

 

[Templeton]

The only force causing the car to decelerarate is the force of friction:
Ff = ma
uFn = ma
umg = ma
ug = a
solve for a, should be ~-6.87 m/s^2
plug in to 2as = Vf^2 - Vo^2
2(-6.87m/s^2)(20m) = -Vo^2
Vo = 16.6 m/s

Hope that helps

 

[notfred]

I'm no physics expert, but wouldn't there be more factors than that? How many skid marks were there? 4? or jsut two? how wide were the tires? Shouldn't skinnier ones take longer to stop?

 

[MajesticMoose]

Originally posted by: notfred
I'm no physics expert, but wouldn't there be more factors than that? How many skid marks were there? 4? or jsut two? how wide were the tires? Shouldn't skinnier ones take longer to stop?

That information is factored into the coefficent of friction IIRC

 

[tontod]

Originally posted by: Templeton
The only force causing the car to decelerarate is the force of friction:
Ff = ma
uFn = ma
umg = ma
ug = a
solve for a, should be ~-6.87 m/s^2
plug in to 2as = Vf^2 - Vo^2
2(-6.87m/s^2)(20m) = -Vo^2
Vo = 16.6 m/s^2

Hope that helps

Did you meant to say Vo = 16.6 m/s ?

 

[Templeton]

Originally posted by: tontod

Originally posted by: Templeton
The only force causing the car to decelerarate is the force of friction:
Ff = ma
uFn = ma
umg = ma
ug = a
solve for a, should be ~-6.87 m/s^2
plug in to 2as = Vf^2 - Vo^2
2(-6.87m/s^2)(20m) = -Vo^2
Vo = 16.6 m/s^2

Hope that helps

Did you meant to say Vo = 16.6 m/s ?

Yes, I did. Thanks

 

[Gibson486]

here is the equation I used

ma-mgu=-ma

u=coeffeienct of friction, kenetic.

I thought that you need the force the car is going, which is ma

Than you need the friction force that is slowing it down, which is mgu, and since it is slowing it down, it should be subtracted.

Then, it should equal -ma because acceleration is negitive due to the car slowing down.

I get 3.43 m/s^2. What am I doing wrong?

 

[Gibson486]

I do not get why friction is the only force acting upon the car. Shouldn't the original Force, which is goimg in the other direction be also in the equation?

Also, I am having trouble with this question

A wet bar of soap slides freely down a ramp 8.0 m long inclined at 6.6°. How long does it take to reach the bottom? Assume u= 0.060.

Here is what I did
Q=theta

F(net)=ma
mgsinQ-umgcosQ=ma
Plug everything in, I get a = .54 m/s^2

To get the time,
I used x(f)=x(i) + v(o)t + .5at^2.
I get 1.72s, but it's wrong! Arggggg!

 

[Gibson486]

ignore the last question I plugged in my numbers wrong

 

[Templeton]

There is no original force, to have a force, you need an acceleration that acts on a mass. The only thing affecting the cars acceleration is the force of friction. Think of it this way, if there were no force of friction, the car would not experience any acceleration. It would keep a constant velocity. During this constant velocity, since the car is not accelerating, there are no forces acting on it. (neglicting wind resistance or other such crap) The only thing that has an affect on the car stopping is friction. Hope this helps.

 

[MacBaine]

Are you ever going to tell me what happened with that other problem with the guy throwing the ball to his friend on the building? You can't solve it... I want to know what your teacher said...

 

[Gibson486]

I didn't even ask him. I doubt it could of been solved either. I look at it again and ask the guy who works in teh physics workshop.

 

[MacBaine]

Originally posted by: Gibson486
I didn't even ask him. I doubt it could of been solved either. I look at it again and ask the guy who works in teh physics workshop.

Yes... I am interested in what your teacher has to say. If he thinks that you can solve it by taking the tan^-1 of the measurements, thus finding the angle of elevation to the tip of the building, he has no idea what he's teaching.

 

[RaynorWolfcastle]

care to fill me in as to what the problem was? sounds interesting...

 

[Gibson486]

3) A person standing 47.5 m away from a 23.m high building wants to throw a ball to a friend. The ball is thrown from a height of 2 m.

a) what is the smallest vertical component velocity of the ball at realease required to reach the top of teh building?

b) What is the least horizontal componenet of velocity for the balll to reach the near edge?

c) How long will it take for the ball to reach the top of the building if it is thrown with that velocity?

 

[Templeton]

I think to solve that problem, you should assume an angle of 45 degrees relative to the horizontal. At 45 degrees, a projectile will always travel the farthest, so the least velocity will be required for it to travel a set distance.

 

[xWeston]

Originally posted by: Gibson486
3) A person standing 47.5 m away from a 23.m high building wants to throw a ball to a friend. The ball is thrown from a height of 2 m.

a) what is the smallest vertical component velocity of the ball at realease required to reach the top of teh building?

b) What is the least horizontal componenet of velocity for the balll to reach the near edge?

c) How long will it take for the ball to reach the top of the building if it is thrown with that velocity?

A) The smallest vertical component would be with the fastest horzontal component, because that will give it the least amount of time in the air, right?

b) the least horizontal component would be as the y component reaches infinity with an angle going to 90 degrees

c) infinity seconds

There has to be more to the problem than what you have given here

 

[ChefJoe]

Well, nobody ever said you were looking for a number solution... you should be able to get the answers as equations with variables left.

 

[Moonbeam]

The smallest vertical component looks to me like the velocity of a ball at impact dropped from 21 meters. Once you know the velocity, you can compute the time it would take to fall. That will be the time it takes to rise too. Knowing the time and the distance 47 M you can solve for the horizontal velocity.

 

[silverpig]

Originally posted by: Moonbeam
The smallest vertical component looks to me like the velocity of a ball at impact dropped from 21 meters. Once you know the velocity, you can compute the time it would take to fall. That will be the time it takes to rise too. Knowing the time and the distance 47 M you can solve for the horizontal velocity.

That is the correct answer for (a).


xWeston has the correct answers for (b) and (c).

 

[silverpig]

Originally posted by: notfred
I'm no physics expert, but wouldn't there be more factors than that? How many skid marks were there? 4? or jsut two? how wide were the tires? Shouldn't skinnier ones take longer to stop?

friction is dependent on only 2 things:

The force between the two surfaces normal to their surface of contact, and the materials.


Surface area makes absolutely no difference in a perfect world situation.

 

[Calin]

DELETED - response to initial problem

 

[Calin]

Originally posted by: Gibson486
3) A person standing 47.5 m away from a 23.m high building wants to throw a ball to a friend. The ball is thrown from a height of 2 m.

a) what is the smallest vertical component velocity of the ball at realease required to reach the top of teh building?

b) What is the least horizontal componenet of velocity for the balll to reach the near edge?

c) How long will it take for the ball to reach the top of the building if it is thrown with that velocity?

a. v^2 = 2*g*deltaH, deltaH = 21m (this is vertical component)
so v=20.3 m/s (vertical speed)

b. If I throw with vertical speed very much, horizontal speed can be very low (the ball stays in air alot of time). But normally you compute speed for the minimum vertical speed.
time in air = v/g = 20.3/9.81 = 2.07s
so orizontal speed is 23 m/s
c. see b.

ATTN: a solution which finds correct minimum involve (in my opinion) differential calculus. This is high school math

Calin

 

[Moonbeam]

t squared =D/.5g time = 2.07 seconds for a ball to fall from or rise to 21 meters. The initial velocity then has to be v= gt=9.8x2.07=20.3M/sec or an average velocity of 10.2M/sec the verticle velocity

Since the ball has to travel the 47.5 meters in the same time it climbs the 21 meters and since it does it in 2.07 seconds v=d/t=47.5/2.07=22.9M/sec for the horozontal velocity.

Is this wrong because I don't understand xWeston's explanation.

 

[silverpig]

Originally posted by: Moonbeam
t squared =D/.5g time = 2.07 seconds for a ball to fall from or rise to 21 meters. The initial velocity then has to be v= gt=9.8x2.07=20.3M/sec or an average velocity of 10.2M/sec the verticle velocity

Since the ball has to travel the 47.5 meters in the same time it climbs the 21 meters and since it does it in 2.07 seconds v=d/t=47.5/2.07=22.9M/sec for the horozontal velocity.

Is this wrong because I don't understand xWeston's explanation.

This is fine if you assume that the teacher wanted your response to (b) to be based on your response to (a).

What xWeston is saying is that you could throw the ball up at 89.999 degrees with a velocity of v = 10000000 so that it will go nearly straight up. This will cause it to be in the air for a very long time, thus allowing your horizontal velocity to be very small. Real world though, v would have to be less than the escape velocity of the earth, so you do have some limit there.