# Find stopping distance from mass velocity & frictional force

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1. Feb 27, 2016

### bchq333

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

A 50,000kg locomotive is travelling 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?

Last edited by a moderator: Feb 28, 2016
2. Feb 27, 2016

### Dr. Courtney

Not too far off.

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

3. Feb 27, 2016

### bchq333

d is the initial velocity times stopping time

4. Feb 27, 2016

### Dr. Courtney

That is what you did, but is it right?

5. Feb 27, 2016

### drvrm

A 50,000kg locomotive is travelling 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/

6. Feb 27, 2016

### bchq333

Average velocity = displacement/time. Wouldn't the displacement be the same as the distance, as it is only travelling in a straight line? therfore 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.

7. Feb 27, 2016

### bchq333

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

Is that correct?

8. Feb 28, 2016

### normal_force

d=vi•t+1/2(at)

9. Feb 28, 2016

### olgerm

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

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

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

Last edited: Feb 28, 2016
10. Feb 28, 2016

### 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, lets 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.

Last edited: Feb 28, 2016
11. Feb 28, 2016

### olgerm

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

12. Feb 28, 2016

### normal_force

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

13. Feb 28, 2016

### normal_force

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.