Question

3. How does kinetic energy affect the stopping distance of a small vehicle compared to a large vehicle?

Answer 1

Stopping distance does not depend on mass at all.

Answer 2

Because work done = change in KE, which will just be equal to the initial KE since final is zero, and work done = Fd
\[\large F d = \frac{ 1 }{2 } mv^2\]where F is the friction force, mu*Fn, and in this case, the normal force would equal to mg
\[\large \mu mg d = \frac{ 1 }{2 } mv^2\]
solve for d \[\large d = \frac{ v^2}{\mu g } \]notice distance only depends on velocity and the road surface friction.

Answer 3

I was wondering if could you have used momentum also

Answer 4

Change in momentum would depend on time, which would mean you'd have to first solve for it

Answer 5

this time we want momentum to be \(p = 0 \), which means the velocity is 0
maybe it is a lot more work and I was actually working it out earlier but I end up doing more than this typical step of doing kinetic energy+potential energy + work = kinetic + potential

Answer 6

the only problem is that we can't account for distance when using momentum.

Answer 7

or maybe we can ?

Answer 8

I'm not sure I'd immediately know how to use momentum in this kind of situation, since I'm so used to solving only collision problems with conservation of momentum... as for stopping distance itself, that's an interesting question, because automobiles tend to disk brakes as the primary decelerating force... the stopping distance due to the road might only be applicable in cases of low drag, bearing friction (not to mention that we need to account for rolling resistance here)... I guess what we're discussing here is more accurately coasting distance...

Answer 9

I'd actually have to disagree with @agent0smith if considering stopping distance as braking distance (what is this "stopping distance" we are speaking of, @josedavid ?)... just look how much distance a big train needs to stop

Answer 10

well it says kinetic energy so ya we go for it, but without that keyword, any moving object such as trucks (at least to me) I think of momentum

Answer 11

If this is a question from a physics class, which it likely is, then it's based on braking distance with wheels locked. I've taught physics, it's a common question.

Answer 12

Momentum in itself has always been a weird concept to me, in particular because momentum is always conserved despite inelastic collisions... I think the problem we run into is having to identify our collision... in this case, wouldn't it actually be conservation of *angular* momentum due to the earth, and the tiny car trying to spin it?

And yes, that makes a lot more sense now @agent0smith ... funny though how they give us these problems and totally ignore modern technology like ABS brakes ;)

Answer 13

Momentum isn't conserved in this case. Unless like you said you consider the earth and car as your system.

Answer 14

There's an external force on the car so its momentum isn't conserved.