Why do I keep getting friction coefficient as a negative?

[Iamconfused123]

Homework Statement

A truck of a mass 5000kg is moving at 10m/s and we turn the engine off, the car stops after 5 seconds, what is the coefficient of friction?

Relevant Equations

mi - a friction coefficient
F=ma, Ffr=mi*m*g

I mean, I know why I am getting that result(because acceleration is negative), and I know that is wrong butI don't understand how do I get the correct answer...

m=5000kg
Vi=10m/s -> initial velocity
Vf=0 -> final velocity
t=5s

mi=?I first calculated acceleration(Vf=Vi+at -> a=-10/5= -2m/s^2. And then Ffr=mi*m*g and since Ftr=ma then ma=mi*m*g -> mi=a/g=-2/10=-0.2

Can someone please help me with this. I've tried similar problem, just with distance crossed but no time given, and got negative coefficient again.

 

[Orodruin]

Your misrake is assuming that the frictional force acts in the direction of motion. It does not. It acts opposite to the direction of motion.

 

[kuruman]

The force of friction is the only force acting on the truck hence it is the net force. If the acceleration is negative the net force must be negative.

 

[kuruman]

Well, yes, but, am I allowed to just add - myself in front of mi(fric. coeff) in formula Ffr=mi*m*g? Because that is not the original formula, would that affect the result if I had to do multiple steps or if I had 2 eq with two unknowns?

Yes you are allowed. The formula gives you the magnitude of the force of friction. You must put the minus sign in front of the magnitude to make a vector pointing in the same direction as the acceleration. If you don't, you will be violating Newton's second law.

 

[Iamconfused123]

Thanks, I deleted that comment. Thought that was stupid. But it just a bit odd for me to just add - myself.
I don't think I ever did that before in any math or physics problem.

 

[kuruman]

I mean I know I am allowed to do that, but what if I was for example writing some type of code, where I'd have to go and manually add/replace all -es or +es? That would be very stupid job to do

"Allow" is a weak word. You are required to do that because, as I already mentioned, you will be violating Newton's second law if you don't match the direction of the net force and the direction of the acceleration.

If you make no mistakes, you will not have to replace anything whether you solve a physics problem or write code.

 

[Iamconfused123]

"Allow" is a weak word. You are required to do that because, as I already mentioned, you will be violating Newton's second law if you don't match the direction of the net force and the direction of the acceleration.

If you make no mistakes, you will not have to replace anything whether you solve a physics problem or write code.

Thank you for helping me.

 

[haruspex]

Homework Statement: A truck of a mass 5000kg is moving at 10m/s and we turn the engine off, the car stops after 5 seconds, what is the coefficient of friction?

Is this question valid? Turning the engine off is not the same as hitting the brakes. The vehicle's momentum can keep the engine turning over for a while.
The coefficient of friction between road and tyres could be much higher than computed from this info.

 

[kuruman]

Is this question valid? Turning the engine off is not the same as hitting the brakes. The vehicle's momentum can keep the engine turning over for a while.
The coefficient of friction between road and tyres could be much higher than computed from this info.

I wouldn't turn off the engine of a 5-ton truck expecting it to stop in 5 s even when it's moving at 36 km/h. With the engine off, there will be at least partial loss of steering and braking ability. In any case, this question obliges one to assume that turning off the engine locks the wheels which is certainly not what happens.

 

[PeroK]

Homework Statement: A truck of a mass 5000kg is moving at 10m/s and we turn the engine off, the car stops after 5 seconds, what is the coefficient of friction?

This question is nonsense and must have been written by someone who a) has never driven a car or truck; and, b) has never studied physics. I wonder if it was AI generated?

The first problem is that, AFAIK, to stop a truck you apply the brakes, rather than switch off the engine.

If you do not apply the brakes, the truck would eventually come to a stop due to a combination air resistance and rolling resistance.

If you apply the brakes normally, the truck would stop quickly due to static friction between the tires and the road. But, unless you are braking at maximum capability, you would not be using the full coefficient of static friction.

If you lock the wheels by braking too hard, and the truck has no anti-locking braking system, then the truck will slide to rest using the full coefficient of kinetic friction between the tires and the road.

 

[haruspex]

If you do not apply the brakes, the truck would eventually come to a stop due to a combination air resistance and rolling resistance.

In neutral, yes, but if the engine is still engaged then you have the wheels cranking the engine. That will be less resistance than brakes can supply, but rather more than rolling resistance.

 

[PeroK]

In neutral, yes, but if the engine is still engaged then you have the wheels cranking the engine. That will be less resistance than brakes can supply, but rather more than rolling resistance.

Okay, so there are internal factors. But, it's still quite fantastic to think that you stop a truck by switching off the engine.

 

[haruspex]

Okay, so there are internal factors. But, it's still quite fantastic to think that you stop a truck by switching off the engine.

It is important what happens in that event. Maybe the ignition failed or you ran out of fuel.

 

[PeroK]

It is important what happens in that event. Maybe the ignition failed or you ran out of fuel.

Or perhaps the truck was going uphill. That would explain the stopping in only 5 seconds.

 

[Orodruin]

Or perhaps the truck was going uphill. That would explain the stopping in only 5 seconds.

I’d say around 9 seconds



(Although that one is probably going faster than 10 m/s …)

 

[gmax137]

As long as we are this far into the weeds... I had a summer job where we sometimes got to drive a small (5 yard) dumptruck around the construction site. One of the more seasoned guys showed me how to turn off the engine on a downhill stretch, count to three, then switch the ignition back on. Boom!! Endless laughs. Just don't do it when El Jefe is nearby.

 

[berkeman]

One of the more seasoned guys showed me how to turn off the engine on a downhill stretch, count to three, then switch the ignition back on. Boom!!

Was the loss of power steering and power brakes any problem?

 

[gmax137]

Was the loss of power steering and power brakes any problem?

No, we were going only 20 or 25 mph, in a straight line. It is hard on the exhaust pipe tho.