Coasting up a driveway with snow - cons. of energy

[totallyclone]

Coasting up a driveway with snow -- cons. of energy

Homework Statement

Your Physics teacher drives an ideal frictionless car of mass 1000kg which coasts along the level road to his home at a constant speed of 48km/h with the engine turned off. It then coats up 40m his sloping driveway and comes to rest at the top, without any braking required. One day he found that his engine had to develop 30 kilowatts of power to maintain a stead speed of 48km/h on the level due to a layer of loose snow on the road. He quickly calculated the new speed required for him to coast up his driveway with the engine switched off at the bottom as usual. He assumed that the snow on the hill would give the same constant retarding force as the snow on the level. What was the new speed?

Homework Equations

E_Ti=E_Tf
E_k=1/2 mv²
E_g=mgh
P=E/T
S=D/T

The Attempt at a Solution

I drew a diagram and attached it.

48km/h=13.3m/s

30kW=30 000w

Without snow
T=D/S
T=40m/13.3m/s
T=3.0s

E_Ti=E_Tf
1/2 mv²=mgh
1/2 v²=gh
v²/2g=h
(13.3)²/2(9.8)=h
h=9m

θ=sin^-1(9/40)
θ=0.23°


With snow
P=E/T
E=PT
E=(30 000W)(3.0s)
E=90 000J

E=1/2mv²
√2E/m=v
v=√2(90 000)/1000
v=13.4m/s

I'm not sure though... so does that mean there's snow on all of it? When there is snow, he has his engine turned on throughout the level but instantaneously turns it off before he goes up the driveway? I feel that the angle of the driveway is too small.

So, if it reaches the top off the driveway, something must be 0? The acceleration would be 0 wouldn't it?

 

[collinsmark]

I think you're making it way too complicated. There's a much simpler way to approach this problem, and none of it involves kinematics.

I'm not sure though... so does that mean there's snow on all of it?

Yes, I believe so. There is snow all along the driveway.

When there is snow, he has his engine turned on throughout the level but instantaneously turns it off before he goes up the driveway?

Yes, that's the way I understand the problem statement. The engine is turned off at the very moment the car starts up the driveway.

I feel that the angle of the driveway is too small.

Don't worry about the angle of the driveway. (See below.)

So, if it reaches the top off the driveway, something must be 0? The acceleration would be 0 wouldn't it?

Well, sure the velocity is zero at the moment the car reaches the top of the driveway. But no, the acceleration is not zero when the car is moving on the driveway. But there's no need to concern ourselves with the car's acceleration. There is an easier way.

[Edit: The car's acceleration is 0 before reaching the driveway. I suppose this is an important fact to know for coming up with an expression for the frictional force. Long before the car reaches the driveway, and before the car changes to a new speed, it is exerting 30 kilowatts of power to maintain a stead speed of 48km/h. Can you find use that to find the force of friction?]

Without Snow:
With no snow all the forces are conservative. You know that all of the car's initial kinetic energy ends up getting converted to gravitational potential energy.

Using this information, you could calculate the driveway's height and even the angle of the driveway as you have attempted. But there's no need to for this problem. All you really need is the change in gravitational potential energy.

With snow:
The gravitational potential energy doesn't change whether there is snow or not.

What does change is the additional work done on the force of friction. How much work is done on the force of friction?

So, how much initial kinetic energy is required to overcome the change in gravitational potential energy plus the work done on the frictional force?

 

[totallyclone]

With snow:
The gravitational potential energy doesn't change whether there is snow or not.

What does change is the additional work done on the force of friction. How much work is done on the force of friction?

So, how much initial kinetic energy is required to overcome the change in gravitational potential energy plus the work done on the frictional force?

In the question, it says "...drives an ideal frictionless car...". So, apparently, there is a force of friction??

 

[collinsmark]

In the question, it says "...drives an ideal frictionless car...". So, apparently, there is a force of friction??

The car doesn't have any internal friction. But there is friction when driving on the snow.

 

[totallyclone]

The car doesn't have any internal friction. But there is friction when driving on the snow.

So there is an applied for by the engine and a force of friction, then they should both be the same values since they oppose each other.

Also, I realized my calculator was in rad when I calculated the angle of the driveway no wonder it was sooo small. I corrected the angle to be 13.0°.

My power formula was kind of wrong too because I wrote it off the top of my head. I think it's better with P=W/T so this is what I came up with to find F_A:

P=W/T
P=F_AΔd/t
P=F_Av
F_A=30 000W/13.3m/s
F_A=2256N

Therefore F_F=2256N

 

[collinsmark]

So there is an applied for by the engine and a force of friction, then they should both be the same values since they oppose each other.

Also, I realized my calculator was in rad when I calculated the angle of the driveway no wonder it was sooo small. I corrected the angle to be 13.0°.

My power formula was kind of wrong too because I wrote it off the top of my head. I think it's better with P=W/T so this is what I came up with to find F_A:

P=W/T
P=F_AΔd/t
P=F_Av
F_A=30 000W/13.3m/s
F_A=2256N

Therefore F_F=2256N

Yes, very nice.

So, how much work is required to oppose this force over the length of the entire driveway?

 

[totallyclone]

Yes, very nice.

So, how much work is required to oppose this force over the length of the entire driveway?

W_NET=E_Kf-E_Ki
W_NET=-1/2 mv² eqn.1

W_NET=F_fΔdcos180°+F_g||Δdcos180°
-1/2 mv²=F_fΔdcos180°+F_g||Δdcos180° subbed eqn.1
v=√-2(F_fΔdcos180°+F_g||Δdcos180°)/m
v=18.9m/s
v=68.0km/h

I think this makes sense that the speed is more than the speed it would take without the snow now that there is friction.

 

[collinsmark]

W_NET=E_Kf-E_Ki
W_NET=-1/2 mv² eqn.1

W_NET=F_fΔdcos180°+F_g||Δdcos180°
-1/2 mv²=F_fΔdcos180°+F_g||Δdcos180° subbed eqn.1
v=√-2(F_fΔdcos180°+F_g||Δdcos180°)/m
v=18.9m/s
v=68.0km/h

I think this makes sense that the speed is more than the speed it would take without the snow now that there is friction.

Your final answer of 68.0 km/h looks correct to me (ignoring any minor rounding errors*). Good job!

*(You might end up with a slightly different answer if you rework the problem keeping more significant figures in your intermediate steps. But the difference is a small one.)

 

[totallyclone]

Your final answer of 68.0 km/h looks correct to me (ignoring any minor rounding errors*). Good job!

*(You might end up with a slightly different answer if you rework the problem keeping more significant figures in your intermediate steps. But the difference is a small one.)

What would you suggest I leave the answers in how many significant figures?

 

[collinsmark]

What would you suggest I leave the answers in how many significant figures?

There's actually certain formal conventions regarding this, but I don't remember what they are. Every time I've learned them I've quickly forgotten. You might want to search the web on the subject.

Here is what I use though:
If my final answer is to have three significant figures (such as 68.0), then I work with at least 4 significant figures in all intermediate steps. Better yet, I use 5 or more significant figures if I'm doing a lot of calculations since rounding errors tend to propagate and get worse as more numbers are added or multiplied together.

The biggest source of rounding error here was the 48 km/h = 13.3 m/s. It's more like 13.333 m/s.

The difference it makes in your final answer is very small. But it does make a difference.

 

[totallyclone]

There's actually certain formal conventions regarding this, but I don't remember what they are. Every time I've learned them I've quickly forgotten. You might want to search the web on the subject.

Here is what I use though:
If my final answer is to have three significant figures (such as 68.0), then I work with at least 4 significant figures in all intermediate steps. Better yet, I use 5 or more significant figures if I'm doing a lot of calculations since rounding errors tend to propagate and get worse as more numbers are added or multiplied together.

The biggest source of rounding error here was the 48 km/h = 13.3 m/s. It's more like 13.333 m/s.

The difference it makes in your final answer is very small. But it does make a difference.

...right. Yeah, I do remember there's a way to determine how many significant figures one should use. For example, if the given is in 3 significant figures, the suggested answer should be more than 3 significant figures, maybe 4 or even 5.

Anyways, thank you so much for your help on this question! I really appreciate it. Good day