Work and kinetic theorem problem using friction

[Qedpower]

Homework Statement

A 2.1x10^3 kg car starts from rest at the top of a driveway that is sloped at an angle of 20 degrees with the horizontal. An average friction force of 4x10^3 N impedes the car's motion so that the car's speed at the bottom of the driveway is 3.8 m/s. What is the length of the driveway?

Homework Equations

mass of car =2.1x10^3 kg
velocity (final)=3.8 m/s
average friction force= 4x10^3
driveway sloped at 20 degrees
kinetic energy theorem=1/2m(vf)^2-1/2m(vi)^2
work formula=fd(cos θ)
w=ke

The Attempt at a Solution

If I log the numbers into the kinetic theorem equation
ke=1/2(2.1x10^30)(3.8m/s)^2
If I log the numbers into the work equation
w=4x10^3d(cos 20)
and ke=w
doesn't 4x10^3d(cos 20)=1/2(2.1x10^30)(3.8m/s)^2?
If I simplify this, the distance is 4.03. Is there anything wrong with my equation?

 

[gneill]

Homework Statement

A 2.1x10^3 kg car starts from rest at the top of a driveway that is sloped at an angle of 20 degrees with the horizontal. An average friction force of 4x10^3 N impedes the car's motion so that the car's speed at the bottom of the driveway is 3.8 m/s. What is the length of the driveway?

Homework Equations

mass of car =2.1x10^3 kg
velocity (final)=3.8 m/s
average friction force= 4x10^3
driveway sloped at 20 degrees
kinetic energy theorem=1/2m(vf)^2-1/2m(vi)^2
work formula=fd(cos θ)
w=ke

The Attempt at a Solution

If I log the numbers into the kinetic theorem equation
ke=1/2(2.1x10^30)(3.8m/s)^2
If I log the numbers into the work equation
w=4x10^3d(cos 20)
and ke=w
doesn't 4x10^3d(cos 20)=1/2(2.1x10^30)(3.8m/s)^2?
If I simplify this, the distance is 4.03. Is there anything wrong with my equation?

Presumably the frictional force will act directly opposing the motion with no relative angle (perhaps a faulty handbrake?). So your "work equation" for the work due to friction is not correct -- there should be no angle involved there.

A bigger problem is that you've left out one source of energy entirely. What's another source of energy when you go downhill?

 

[Qedpower]

So have I left out gravity? What do you mean when you said that no angle should be involved?

 

[gneill]

So have I left out gravity? What do you mean when you said that no angle should be involved?

Yes, you have left out gravity. The friction force has no angle w.r.t. the motion.

 

[Qedpower]

I have figured out that mg=average frictionxdistance, is this right?

 

[gneill]

I have figured out that mg=average frictionxdistance, is this right?

No, mg is a force (Units of Newtons). Frictional force x distance yields work (units of Joules). Perhaps you meant mgΔh (change in gravitational PE)? Note that only some of the PE gets converted to KE, and some is dissipated due to friction.

Write an energy balance equation for all the energies involved. For a given distance traveled along the driveway, what's the change in gravitational PE? How much energy is lost to friction? Where does the rest go?

 

[Qedpower]

I think I've got it. So there are two forces. The downward motion that is helped by gravity and the car's velocity (call it f1) and the average friction (4000). I will be using the formula distance=1/2at^2. I need to find another meaning for the time and acceleration. So f1= mg cos 20 degrees so f1-4000=force without friction (f2). If f1=ma, and a=f1/m,then velocity=f1time/m. This means that t=mv/f1. So if I log it in, the formula is 1/2mv^2/f1. If I plug it in and calculate it, the answer is 5(rounded). The book's answer is 5.1, so I'm not sure if I am correct or not.

 

[gneill]

I think I've got it. So there are two forces. The downward motion that is helped by gravity and the car's velocity (call it f1) and the average friction (4000). I will be using the formula distance=1/2at^2. I need to find another meaning for the time and acceleration.

Not sure why you'd wand to use (1/2)at₂; wouldn't it be more straightforward to employ the work / energy approach?

So f1= mg cos 20 degrees so f1-4000=force without friction (f2).

I don't see how the above can be true. mg cos 20 degrees would be the surface normal force, not a downslope force. And while subtracting the frictional force from the downslope force due to gravity would yield the net downlope force, you never make use of this later.

If f1=ma, and a=f1/m,then velocity=f1time/m. This means that t=mv/f1. So if I log it in, the formula is 1/2mv^2/f1. If I plug it in and calculate it, the answer is 5(rounded). The book's answer is 5.1, so I'm not sure if I am correct or not.

Your f1 is not the net downslope force acting on the car, so f1=ma will not work.

The only force you should have to deal with is the frictional force so that you can write an expression for the work done by friction over the distance that the car travels (along the driveway). Other than that you have the final kinetic energy of the car and the change in gravitational potential energy due to the car's drop in height over its path. All these ENERGY components are related (think conservation of energy).

 

[gneill]

By the way, it appears to me that the book's answer of d = 5.1m is not accurate for the given parameters of this problem. I suspect that somewhere along the line someone has revised the "given" values in order to give the problem a makeover, but either forgot to update the answer key to reflect the change or miscalculated the new result themselves; I see the result being closer to 5m than 5.1m.