Acceleration and friction force problem

[AlexPilk]

Homework Statement

Mass of the bus is 20 tons. It accelerates to 10 m/s (from initial speed of 0) in 50 meters. Find the coefficient of friction (u) if the force, that pushes the bus = 14*10^3 N

m=20t=20000kg
S=50m
v=10m/s
F=14kN=14000N

The Attempt at a Solution

So we can find out what the acceleration is using
a=2S/v^2
And it will be equal to 1 m/s^2
But, the total force F(total)=F-F(friction)=m*a=14000-F(friction)=20000
So it turns out Force of friction = -4kN. How can this be, what did I do wrong?

 

[Chestermiller]

I agree with you. It doesn't seem to make sense.

Chet

 

[Orodruin]

I also agree. Another way of looking at it would be to look at work done by the pushing force which would equal 700 kJ. The kinetic energy of the bus moving at 10 m/s would be 1 MJ so the same problem manifests itself in that consideration too.

 

[CWatters]

So we can find out what the acceleration is using a=2S/v^2

The SUVAT equation is..
V^2 = U^2 +2as
U=0
so
a = V^2/2s

but despite that you still get a=1m/s^2

I thought perhaps they meant imperial (US short) Tons rather than metric but that's about 18,100kg and you still get a negative answer.

A strong tail wind is the only thing I can think of.

The friction with the ground is actually what pushes the bus so if the "force that pushes the bus is 14,000N" then the friction force with the ground must also 14,000N. But in that case the bus won't accelerate as fast as stated. I agree it makes no sense.

It's not unknown for teachers/examiners to copy old questions and change the values without checking it all hangs together.

 

[HalfLostAndSearching]

Your approach to finding the coefficient of friction (u) is correct, but there are a few errors in your calculations.

First, the mass of the bus should be converted to kilograms, not tons. This means that the mass of the bus is 20000 kg.

Next, the equation you used to find the acceleration is incorrect. It should be a=v^2/2S, which gives an acceleration of 1 m/s^2.

Finally, in your calculation of the total force, you forgot to include the force of friction. It should be: F(total)=F-F(friction)=ma=14000-(-F(friction))=20000. This gives a force of friction of 6 kN, not -4 kN.

Using the correct values in your equation for the coefficient of friction, we get:
u=F(friction)/F(normal)=6kN/20000N=0.0003

Therefore, the coefficient of friction for the bus is 0.0003.