In Problem 2, what is the speed of the car at (a) point A,...

In Problem 2, what is the speed of the car at (a) point $A$, (b) point $B$, and $(\mathrm{c})$ point $C ?(\mathrm{~d})$ How high will the car go on the last hill, which is too high for it to cross? (e) If we substitute a second car with twice the mass, what then are the answers to (a) through (d)?

In Problem 2, what is the speed of the car at (a) point $A$,
(b) point $B$, and $(\mathrm{c})$ point $C ?(\mathrm{~d})$ How high will the car go on the last hill, which is too high for it to cross? (e) If we substitute a second car with twice the mass, what then are the answers to (a) through (d)?

Key Concepts

Mechanical Energy Conservation

This concept states that in a closed system with only conservative forces acting, the total mechanical energy (sum of kinetic and potential energies) remains constant. It is used to analyze situations where gravitational potential energy is converted into kinetic energy and vice versa, such as a car moving along hills or valleys.

Kinetic Energy

Kinetic energy is the energy of motion, calculated by the formula (1/2)mv², where m represents mass and v the speed. It is a key component when determining the speed of a moving object as energy transitions occur between motion and elevation changes.

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field, typically expressed as mgh, where m is mass, g is the acceleration due to gravity, and h is the height above a reference level. This concept is essential in evaluating how high an object will move on an incline during energy conversion processes.

Mass Independence in Conservative Systems

In systems with only conservative forces, like gravitational forces, the mass of the objects often cancels out in the energy conservation equations. This leads to the result that factors such as speed and maximum height reached are independent of mass, which is particularly relevant when comparing outcomes for objects of different masses.

Transcript

00:01 All right, conservation of energy problem here.

00:04 So in all cases, you have the basic principle that initial energy, e initial, is equal to final energy, e -final.

00:14 So for part a, that means initial kinetic energy plus initial potential energy, k -0 -e -not, will be equal to kinetic energy plus potential energy at one at air, excuse me.

00:32 So we know the potential energy is the same at u -not and ua.

00:40 So since they're at the same height, so kinetic energy will be the same at zero and air.

00:49 Therefore, you have, so kinetic energy equals to, is as equal for the two different places, meaning that initial velocity will be the same as velocity at a, which is 17 meters per second.

01:09 In part b, you have k0 plus u0 is equal to kb plus ub.

01:19 And so here, and so let's write the amount.

01:22 Here one half mv not square that's k0 uh u not is m g h being the initial height and kb is one half mv b squared that's what we have to find plus u b that's m g times this time h over 2 that's how much the height has changed so the ms cancel in every term we get rid of the ms we multiply everything both sides by two.

01:57 And what we get is that vb will be equal to the square root of v .0 .2.

02:05 We're just rearranging some terms here at plus 2gh.

02:09 And so v0 squared is 17 meters per second, square that, whoops, square that.

02:18 G is 9 .8 meters per second squared and then h is 42 meters.

02:25 This gives you vb, velocity of b of 26 .5 meters per second.

02:34 In part c, you set k0 plus u0 equal to kc plus uc and really the only thing that changes is, of course vc will be different, because uc is different...

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