Distance traveled by train using forces and kinetic friction

[Mg53]

Homework Statement

A train has a mass of 5 million kilograms and is traveling south at 22 m/s. If the train driver has to initiate a panic stop, how far will the train go before it comes to a halt. Use 0.08 as the coefficient of kinetic friction.

m=5*10⁶ kg
v=-22 m/s
μ-kinetic friction=0.08
g= 9.8 m/s²

Homework Equations

Horizontal surface: F(normal)=F(weight)
F-weight=mg
F-kinetic = (F-normal)*(μ-kinetic friction)
v= sqrt(2*(F-net/m)*x) → x= v²/2*(F-net/m) I'm not sure if it's F(net)/m or g that's supposed to used. My book has both written in it.
F(net)=F(normal)-F(kinetic)
** where F(net)/m comes from the equation F(net)=ma

The Attempt at a Solution

F(weight): (5*10⁶)(9.8) = 4.9*10⁷ N
F(kinetic): (4.9*10⁷)(0.08)= 3.9*10⁶ N
F(net): (4.9*10⁷)-(3.9*10⁶)= 4.5*10⁷ N

F(net)/m: (4.5*10⁷)/(5*10⁶) = a = 9 m/s²

x=(-22^2)/(2*9)
= 54 m

I've tried this problem multiple times but I keep getting the wrong answer. 54 meters doesn't make sense but none of my other answers have either. I think the problem is somewhere in the F(net)/m step though. Is it suppose to be a instead where a=9.8? Any help would be appreciated!

 

[tiny-tim]

Welcome to PF!

Hi Mg53! Welcome to PF!

(try using the X² button just above the Reply box )

F(net): (4.9*10^7)-(3.9*10^6)= 4.5*10^7 N

i don't understand this …

you seem to be subtracting the horiztonal friction from the vertical weight

i] you can't do that

ii] why did you want to?

just do work done = change in kinetic energy ​

(and don't waste space writing in the m, it makes no difference!)

 

[Mg53]

Hi, thanks for the help :)

The reason I did that was because I thought the net force was equal to the normal force minus the force of kinetic friction. Like I said though, I had a feeling that's where I was messing up in the problem but I didn't really know what else to do.

Okay, I'm still a little confused by one part of that.

Work done=change in kinetic energy
=(1/2)(5*10⁶)(22²)
= 1.21*10⁹

I can use that to find the distance using W=Fx but what F do I use? The force of kinetic friction?

I did that and got x=310 m which I guess might make sense but I'm still not sure if that's right because I don't have an answer key :/

 

[tiny-tim]

Hi Mg53!

The reason I did that was because I thought the net force was equal to the normal force minus the force of kinetic friction.

you're completely confused

the net vertical force is zero (because the vertical acceleration is zero)

the net horizontal force is the friction (because it's the only horizontal force)​

… I can use that to find the distance using W=Fx but what F do I use? The force of kinetic friction?

yes of course

(i'm going to bed … goodnight! :zzz:)

 

[Nickie]

Firstly, it is important to clarify that the equation F(net)/m is not the same as a. F(net)/m is the net force divided by the mass, while a is the acceleration due to gravity. In this problem, we are dealing with a horizontal surface, so the acceleration due to gravity is not relevant and should not be used in the calculations.

Next, let's review the steps you have taken to solve this problem. Your calculation for the weight of the train (F(weight)) is correct. However, your calculation for the kinetic friction force (F(kinetic)) is incorrect. The correct calculation should be (4.9*107)*(0.08) = 3.92*106 N. This is because the kinetic friction force is equal to the coefficient of kinetic friction (μ) multiplied by the normal force, which in this case is equal to the weight of the train.

Now, let's calculate the net force acting on the train. We know that the only forces acting on the train are the weight (F(weight)) and the kinetic friction force (F(kinetic)). Therefore, the net force is equal to F(weight) - F(kinetic), which gives us (4.9*107) - (3.92*106) = 4.51*107 N.

Finally, we can use the equation x = v^2/(2*a) to find the distance traveled by the train before coming to a halt. Plugging in the values, we get x = (-22)^2/(2*(4.51*107)) = 53.3 meters.

Therefore, the train will travel approximately 53.3 meters before coming to a halt when the driver initiates a panic stop. It is important to double check your calculations and make sure you are using the correct equations and values. I hope this helps clarify the solution to this problem.