Find stopping distance from mass velocity & frictional force

[bchq333]

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >[/color]

A 50,000kg locomotive is traveling at 10m/s on a level track when the engines and brakes both fail. If there is a frictional force of 142N acting to slow it down, how far will it roll before it stops?

a=F/m
t=v/a
d=vt

a=142N/50,000kg
a=2.84*10^-3

t= 10/2.84*10^-3
t=3521 sec

d=10*3521
d=35211 metres

Stopping distance = 35.21 km? This sounds way too long. Can anyone point me in the right direction?

 

[Dr. Courtney]

Not too far off.

Is d the initial velocity times stopping time or average velocity times stopping time?

 

[bchq333]

d is the initial velocity times stopping time

 

[Dr. Courtney]

That is what you did, but is it right?

 

[drvrm]

A 50,000kg locomotive is traveling at 10m/s on a level track when the engines and brakes both fail. If there is a frictional force of 142N acting to slow it down, how far will it roll before it stops?
a=F/m
t=v/a
d=vt
something wrong here ? check the third equation you are using
Reference https://www.physicsforums.com/threa...e-from-mass-velocity-frictional-force.859569/

 

[bchq333]

Average velocity = displacement/time. Wouldn't the displacement be the same as the distance, as it is only traveling in a straight line? therefore yielding the same answer? Is there another way I can work out distance from what I have? Is what I have for time even correct? 3521 seconds or 58.6 minutes sounds too long as well.

 

[bchq333]

d=1/2*v*t
d=1/2*10*3521
d=17605.65 metres

Is that correct?

 

[normal_force]

d=1/2*v*t
d=1/2*10*3521
d=17605.65 metres

Is that correct?

d=vi•t+1/2(at)

 

[olgerm]

$d=\frac{a \cdot t^2}{2}+v_0 \cdot t$
$a=-\frac{F}{m}$
$t=-\frac{v_0}{a}$

so $d=\frac{v_0^2 \cdot m}{2 \cdot F}$

$d=\frac{10^2 \cdot 50000}{142}≈35211.2676$

 

[normal_force]

Now, this is a good equation to work off of, so I just want to say the KE is just 2,500,000Joules, KE=1/2(50,000kg)(10m/s)^2 that is a lot of work.
We could assume working against a ƒ of 142Newtons, so I'll treat it in the same fashion as work against gravity, or work done against gravitational forces but we are working against a set force of ƒ. I could just take 2,500,000Joules/142N's of ƒ and get 17605.63 meters to slow to zero m/s...just some work-energy basics, as there is a force, a opposing force, we could use that but we would refer to is it as negative force doing negative work.

However, let's pretend we have µ, a "mu" or a "moo" of 0.000289 and we don't know our Ffric, thus our ƒ is µ*m•g, so (.000289)(50000kg)(9.81m/s^2)= 142, I found this by taking the Fgrav/Force norm of 50000kg • 9.81m/s...which is 490500 Newtons, take 142N Friction force / 490500 Newtons of Fgrav/Fnorm(note that Fgrav and Fnorm are the same by Newtons 3rd law), we can find our µ or friction coefficient.

We take 142N's Ffric/ 50000kg's to 0.00284 m/s^2 deceleration...

now, find time we do t=Vf-Vi/a, so a=(0m/s)-(10m/s)/0.000289m/s^2 so we get 3521.12 Seconds.

to find displacement, we can take this data to get our displacement or stopping distance, d=vi•t=1/2(at)

d=(-10)(3521.1267s)+1/2(0.00284)(3521.1267s^2), we can get our final displacement 17605.63 but we would write that as -17605.63 as the -10m/s we are working against the opposing velocity of train as we are using Ffric to stop the train, a retarding or opposing force. It is written because that is the final DISPLACEMENT of the train with the force acting on it, that is why it is negative.

 

[olgerm]

$d=\frac{10^2 \cdot 50000}{142}≈35211.2676$

$d=\frac{10^2 \cdot 50000}{2\cdot142}≈17605.6338$
I calculated wrongly.

 

[normal_force]

$d=\frac{10^2 \cdot 50000}{2\cdot142}≈17605.6338$
I calculated wrongly.

hehe, I was about to say dude
Now there is another, if "better" way to calculate this using another equation.

 

[normal_force]

Not too far off.

Is d the initial velocity times stopping time or average velocity times stopping time?

It would be much better if he knew some equations, specifically just displacement but he could check his work entirely with the equation of Vf^2=Vi^2 + 2ad, he would be finding the net force exerted on the train, so he could check his work with some dynamic problems.