Kinetic energy & Conservation of energy

[Taylan]

Homework Statement

You are driving with your car (of total mass: 1.2tonnes) with a speed of v=50km/h, until you see an obstacle.

a) What is the kinetic energy of the car?

b) When you start to brake, there is still 15m until the obstacle. What must be the size of the friction coefficient (µ), so that you can stop just in front of the obstacle? and what must be the size of the breaking force, Fb? Assume that the friction coefficient and the braking force are constant and use the idea of energy conversation

Homework Equations

KE = 0.5mv²
W= F.d.cos(x)

The Attempt at a Solution

a) KE= 0.5x(1.2x10³kg)(125/9m/s)² = 115.7kJ

b) So I know that since energy (in this case kinetic energy) is never lost, the 115.7kJ must be converted into thermal energy created by the friction. I also know that µR= Friction. So ı guess 15m(µR+ Fb) = 115.7kJ but ı have two unknowns here; µ and Fb (braking force). So I couldn't understand how to cope with this

 

[kuruman]

Assume that the car does not have ABS (anti-lock bake system) and that the brakes "lock". Then friction is the only force that retards the car.

 

[Taylan]

Do you mean I should assume that friction force is same with this braking force that is mentioned?
Which would give out:

115700J = Fb x 15m

Fb= 115700J / 15m = 7713J

and (µR) *15m = 7713J

so µ= 7713J /(15m x 1.2x10^3kgx9.81m/s^2) = 0.0437

does that look fine?

 

[CWatters]

So ı guess 15m(µR+ Fb) = 115.7kJ but ı have two unknowns here; µ and Fb (braking force). So I couldn't understand how to cope with this

There aren't _two_ forces between the car and road.

If in doubt draw a free body diagram.

 

[kuruman]

Do you mean I should assume that friction force is same with this braking force that is mentioned?
Which would give out:

115700J = Fb x 15m

Fb= 115700J / 15m = 7713J

and (µR) *15m = 7713J

so µ= 7713J /(15m x 1.2x10^3kgx9.81m/s^2) = 0.0437

does that look fine?

Looks fine.

There aren't _two_ forces between the car and road.

I think OP initially assumed that there is friction at the brake pads and at the road. See post #2.

 

[Taylan]

thanks a lot!

 

[CWatters]

Looks fine.

I think OP initially assumed that there is friction at the brake pads and at the road. See post #2.

There is but only one is an external force that acts on the car. Hence the suggestion to draw a fbd.

 

[CWatters]

Think I was writing #4 when #3 was posted.

 

[kuruman]

There is but only one is an external force that acts on the car. Hence the suggestion to draw a fbd.

Sure, but if the initial KE is divided into heat by friction generated at the brake pads and the tires, one has to know what fraction of the KE goes where in order to find μ.

 

[gneill]

Do you mean I should assume that friction force is same with this braking force that is mentioned?
Which would give out:

115700J = Fb x 15m

Fb= 115700J / 15m = 7713J

Watch your units. Forces don't have units of Joules.

and (µR) *15m = 7713J

so µ= 7713J /(15m x 1.2x10^3kgx9.81m/s^2) = 0.0437

does that look fine?

Nope. You're setting an energy equal to a force in the above (see above correction about units). Your (µR) *15m yields an energy (Joules), and you previously calculated the number 7713 as the braking force.

 

[CWatters]

Sure, but if the initial KE is divided into heat by friction generated at the brake pads and the tires, one has to know what fraction of the KE goes where in order to find μ.

We might be at cross purposes but the required external friction force is the same regardless of where the energy is dissipated.

There are three cases..

1) No skidding. The friction force between tyres and road is static friction. No energy is dissipated in the road/tyres because the relarive displacement is zero. Friction between brake pads and discs is kinetic friction so all the energy is dissipated in the brakes.

2) Wheels locked. The friction force between tyres and road is kinetic friction. So all of the energy is dissipated in the tyres/road. The friction between brake pads is static friction so no energy is dissipated in the brakes.

3) Skidding but wheels not locked. Kinetic friction in both cases. Energy dissipated in both places. Need more info to apportion energy to either place.

In all three cases the required friction force between tyres and road and the coefficient of friction is the same.

 

[kuruman]

Looks fine.

I take it back. When I looked at post #3, I checked to see if the denominator is correct but I didn't check the numerator. Good catch @gneill!

 

[Taylan]

oh yes , thanks for correcting. So once again

Fb= 7713N
µR= Fb
µ= 7713N/ (1.2x10^3 kg x 9.81 m/s2) = 0.655

I guess ı got it now?

 

[kuruman]

We might be at cross purposes but the required external friction force is the same regardless of where the energy is dissipated.

There are three cases..

1) No skidding. The friction force between tyres and road is static friction. No energy is dissipated in the road/tyres because the relarive displacement is zero. Friction between brake pads and discs is kinetic friction so all the energy is dissipated in the brakes.

2) Wheels locked. The friction force between tyres and road is kinetic friction. So all of the energy is dissipated in the tyres/road. The friction between brake pads is static friction so no energy is dissipated in the brakes.

3) Skidding but wheels not locked. Kinetic friction in both cases. Energy dissipated in both places. Need more info to apportion energy to either place.

In all three cases the required friction force between tyres and road and the coefficient of friction is the same.

Yes of course. The net horizontal force required to stop the car over the given distance $d$ must come from friction alone and will be $f=KE/d$ regardless of how or where its components are generated. That much is guaranteed by the work-energy theorem. One can go further and define an "effective" coefficient of kinetic friction, $\mu_{eff}=\frac{KE}{mgd}$ which is what (I think) was intended to be the answer to this problem.

 

[haruspex]

Assume that the car does not have ABS (anti-lock bake system) and that the brakes "lock". Then friction is the only force that retards the car.

This is irrelevant. ABS still depends on friction to stop the car; it just ensures maximum use is made of static friction, and is no different from the behaviour of an ideal driver. The question does not specify kinetic friction.

 

[kuruman]

This is irrelevant. ABS still depends on friction to stop the car; it just ensures maximum use is made of static friction, and is no different from the behaviour of an ideal driver. The question does not specify kinetic friction.

Right. See #14.

 

[haruspex]

Right. See #14.

Ok, but

Skidding but wheels not locked.

I don't think that can happen, except transiently, on a given wheel at least. One of the two will win control of the wheel.

 

[CWatters]

Ok, but

I don't think that can happen, except transiently, on a given wheel at least. One of the two will win control of the wheel.

Agreed.

If you watch the likes of Lewis Hamilton they sometimes accidentally lock the wheel(s) but manage to then back off just enough so the wheel starts rotating again but not at the road speed.