How Much Force is Needed?-Part Two.

[Medgirl314]

Homework Statement

A 1200 car traveling at 110 km/h comes to a stop in 8.0 seconds. Assume that the acceleration is constant. What force is required to stop the car?

Homework Equations

F=ma
a=Δv/Δt

The Attempt at a Solution

First I converted so that I could give the answer in Newtons.
v=30m/s
m stays the same.
T=8 seconds
a=Δv/Δt
a= 3.75 m/s^2

At this point, I think either my unit conversion or my equation was wrong,as 3.75 m/s^2 seems somewhat slower than usual for a physics problem involving a car. But continuing with the problem, I get 450 Newtons. I'm new to Newtons, so I don't know if this is a reasonable answer. Could I please get a confirmation?

Thanks!

 

[lendav_rott]

What is the mass of the car? Is it 1200kg?
Your answer would be 4500N in that case.

Note that this is not work done all the way to stop the car. It is an amount of force required to slow the car's velocity by 3.75m/s in every passing second.

If you want to calculate all the work that is required to stop the car you would have go further: there are 2 ways to go about it, if you are interested.

 

[Medgirl314]

The mass of the car was 1500 kg.

Oh, okay. So my answer is correct? Would you please help me wrap my mind around the idea that the sheep stops faster than the car? Is it just because the sheep was moving slower, so it was easier to stop?

Yes, I would love to try calculating the work! I haven't gone nearly that far in my physics class yet,so it may require slowly walking through it. Thanks!

 

[lendav_rott]

Okay, 1 thing if the mass of the car is 1500kg then the deceleration doesn't change, however the force does.

F= ma = mΔv/Δt = 1500*(-30)/8 = -5625N why a minus? Δv means (end velocity) - (initial velocity). What you would like to end up with is 0m/s, therefore it is 0 - 30 = -30m/s and it's where the minus comes in. F is a vector - any vector composes of a magnitude(length) , direction and destination.
Since the positive direction is pointed towards where the car moves, the counteracting force would have to be directed with the opposite direction hence -5625N.

You found the acceleration which is -3.75m/s² and you also have the time, which is 8 seconds. The distance the car travels during deceleration and coming to a halt is
s = at²/2 = -3.75*64/2 = -120m .
Now, you can say it just is 120m, because the car is still moving forwards not backwards, but don't flip the script. If you get a minus don't remove it.

The total work done by any force is the (magnitude of the force) * (the distance it is being applied). A = F * s or if you mark distance as d it would A = F * d. I am from Europe, I mark work done as A, you can mark it as W or anything you want. Although always mark down which letter means what so there would be no confusion.
A = -5625N * (-120)m = 675 000 N*m or J (Joules)

I said there are 2 ways. You could also calculate the kinetic energy of the car at the initital velocity which is E_k = mv²/2 = 1500*900/2 = 675 000 J
What does this mean? During the deceleration, the counteracting force would have completely cancel the car's initial kinetic energy for it to stop. If it has no kinetic energy, therefore it isn't in motion. It is the "Conservation of energy". Energy cannot appear out of nowhere and evaporate into thin air. It is always changing between different states.

As you can see from the calculations - it depends on the mass of the object and its velocity determining how hard it is to stop. Should you have a sheep capable of moving at like 100km/h then certainly, it would be difficult to stop it, but it doesn't have a mass nearly as high as a car's.

 

[Medgirl314]

That's really neat, thank you! I really appreciate the explanation, you are so knowledgeable in physics. I think I missed the part where you said whether my answer was correct, would you mind confirming?

Thanks again!

 

[lendav_rott]

Your initial post is lacking of information, therefore I assumed the car's mass to be 1200kg for some reason. Since you pointed out that it is 1500kg I fixed the calculations as well. Your calculations for accelerations and such are correct, however the rest.
I am confused actually how you get the 450N in the first post.

You are welcome

 

[Medgirl314]

Oh, I'm sorry! I confused this post with a completely different one involving a sheep! Yes, the mass of the car in this case was 1200 kg. I hope you may have gotten a laugh though. XD

 

[lendav_rott]

Replace 1500kg with 1200kg in my above calculations and you will arrive at slightly different numbers, however the bottomline is the same.

 

[Medgirl314]

Okay, got it. So in this case, would my answer be correct? It almost seems too low.

 

[lendav_rott]

That depends on what your answer is. Also, get into the habit of sharing your work with the rest of the class. That way help will arrive sooner rather than later.
If you mean the 450N you got in the first post - for a 1500kg car to be decelerated by 3.75m/s every second, it is waaaay to weak. Always think about roughly what your answer would look like. If the force acting on car is 450N, it is something like the car smashing into a 45kg weight, maybe a small stray animal, if you want to get maniacal about it. Is it enough to slow the car down by that much? I don't think it is. Therefore, yes, it is too low.

 

[Medgirl314]

Oops, sorry about that! I was experimenting, I wasn't sure if it was annoying to post every step or not. Looking at my scratch paper, I left out a zero. I got 4500 Newtons. Oops!

 

[Medgirl314]

4500 is correct, correct? (English is odd.)