I still don't understand why the initial velocity of the moving car is equal to the final velocity of both vehicles together? Could you explain this please?

Hi dmorochnick, thanks for the question. Earlier in the video I mentioned that the coordinate system was chosen so that down is positive. While the convention is that if one says nothing about the coordinate system, the assumption is that up is positive, is OK to explicitly change that. The choice is just personal preference. Where it's convenient, I sometimes prefer to make down positive to avoid substituting many negative values into the formula since I find so many negatives, and subtracting negatives, to be confusing. If you wish to have up as positive, it's totally OK to do so. You would arrive at the same answer (not $4.3 \textrm{ m}$), since keep in mind that the displacement past the window would also have to be negative since it's downward (which in your coordinate system is the negative direction if up is positive).

Hi nalakija, thanks for the comment. Could you please be more specific? I'm seeing editions labeled as "7th Edition", "6th Edition" and "Global Edition" in the menu bar. Where are you seeing them labelled as one?

Hi, how do you distinguish whether to use KEf + PEf = KEi + PEi on a spring question like this, rather than using Wext = DeltaPE in a question like #35?
Thank you

Thanks for the question. The formula $KE_f + PE_f = KE_i + PE_i$ is a law, which means it's always valid. This is an algebraic way of saying energy is conserved after a movement of some kind. As a law, it could also be used in #35 instead of $W_{ext} = \Delta PE$. Doing so would have looked like this: $0 + mgy_2 = 0 + mgy_1 + \dfrac{1}{2}k(y_2 - y_1)^2$, where the zero's are substituting for kinetic energy, of which there's none since in #35 the block isn't moving (or the movement is so slow as to be negligible), and $(y_2 - y_1)$ is the amount the spring was stretched. Using $W_{ext} = \Delta PE$ is just a convenience since it reduces a couple of algebra steps. When in doubt, use conservation of energy.

Thanks for the question. I'm away from my recording equipment at the moment, but I'll try to give some highlights here. We start with saying the total forces directed up equal the total forces directed down, then I make a substitution for each up force with the archimedes principle that the buoyant force is the weight of fluid displaced, which is $F = \rho V g$, and the down forces are the usual weight formula $F = mg$. Then, in green, the volume of the wood and lead is expressed in terms of their density and mass. This is in turn substituted (using red) for each volume factor in the up force terms. The specific gravity of wood replaces the ratio of the density of water to wood. Then comes a confusing line where the mass of the wood is shown moving to the left, but actually ends up on the right. The thing to notice with that line is that we're collecting like terms. We're placing the wood terms together on one side, and the lead terms on the other side. The mass of lead and mass of wood are factored out on their respective sides, leaving us with some bracket multiplied by a mass on each side. We're interested in knowing the mass of lead, so we divide both sides by the bracket by which the lead mass is multiplied. Then we plug and chug. Hopefully this video commentary is a bit helpful...

Thanks for the question. "initial" and "final" is a bit confusing in this questions since there are more than one time periods to consider, each with their own "initial" and "final" moment. For part a), the "final" moment is when the car has reached $55 \textrm{ km/h}$, making $\omega_f = 38.194 \textrm{ rad/s}$. However, for part b), the initial moment is when the car is at $55 \textrm{ km/h}$, so for part b) $\omega_i = 38.194 \textrm{ rad/s}$. The initial moment of part b) is the final moment of part a).

I can see where you're coming from, and technically it would be more correct for me to put a subscript on the variables to assign them to parts of the question, such as for the angular speed for part a) written as $\omega_{iA}$ and $\omega_{fA}$, and then say $\omega_{fA} = \omega_{iB}$, and so on, but I think that also creates it's own source of confusion by presenting so many more subscripts.

Why is this collision deemed "inelastic" in terms of the treatment of kinetic energy conservation. Seems based on the diagram, the objects are not sticking together, which would suggest its inelastic.

Thanks for the question. The term "inelastic" means there is less total kinetic energy after a collision compared with before a collision. Since that's the case here, we can call it inelastic. It is not "completely inelastic", that's true. A completely inelastic collision involves the two particles sticking together, and this type of collision results in the maximum loss in kinetic energy that's possible while still conserving momentum. Seeing two particles bounce apart means either the collision is "inelastic" (but not "completely inelastic"), or maybe the collision is "perfectly elastic", in which case the total kinetic energy after collision is the same as it was before. In this case, the collision is a regular inelastic collision since some kinetic energy was lost, but not the maximum loss that would have otherwise been possible had they stuck together.

Could you also approach the problem by first solving for final velocity by using V=a*t, and then plugging this velocity into v^2=vo^2+2AD to solve for displacement?

Yes you can! Substituting $v=at$ info $v^2 = v_o^2 + 2aD$ gives $a^2t^2 = v_o^2 + 2aD$, which, with $v_o = 0$ rearranges to $D = \dfrac{1}{2}at^2$ after you divide both sides by $2a$. $D = \dfrac{1}{2}at^2$ is the formula I used in the video. Good work!

Thanks for the question. One thing to understand about potential energy is that you can never know how much there is. :) There's no such thing as an "absolute" potential energy. It's possible only to tell what is the change in potential energy. Sometimes we speak as though there is an absolute potential energy, such as "the book is 3m above the ground, so it has $mg(3 \textrm{ m})$ of potential energy, but don't be fooled by that as it's just a concise (maybe lazy?) way of speaking, when in fact it means the difference in potential energy of the book with respect to the ground is $mg(3 \textrm{ m})$. With that in mind, the skier has a change in potential energy resulting from their drop in altitude of $230 \textrm{ m}$, and this change in potential energy is turned into other forms of energy, namely kinetic energy and heat. Subtracting away the kinetic energy gained from the loss of potential energy, we're left with how much of that potential energy loss was turned into heat.

To put it in one sentence, the "potential energy" term you see in the final calculation is not the "potential energy at the end", but rather, it is the change in potential energy when the skier descends.

Hi leopinzon18, for formulas being used are $\tau = Fl_{\bot}$ which says torque equals force times the perpendicular component of the lever arm, and the other formula is that $\Sigma \tau = 0$ for static equilibrium, which is also to say $\Sigma \tau_{ccw} = \Sigma \tau_{cw}$ meaning the counter-clockwise torques equal the clockwise torques. If you have a more specific question, just let me know and I'll see if I can answer it.

Hello!
I am still confused as to why the diagrams for a) and b) are the same. I don't understand how in a) the box has static friction, but b) has kinetic friction and their diagrams are the same. Shouldn't b) have a velocity vector pointing downward?
Thank you so much for your time!

Hi schamorg, thanks for asking this question. Understanding free body diagrams is really important, so it's good your digging into the topic. A free body diagram does not include a velocity vector. I sketched one in for part "c", but that was just to emphasize which direction the box is moving in order to explain why the force arrows are pointing the way they are. Notice that the velocity vector does not begin on the center of the box, since I didn't want to make is seem like another force vector. Force vectors are always written with their beginning a the center of mass of the object (usually the geometric center... just ignore this point if it's confusing). Technically, velocity vectors do not belong on free body diagrams. OK, so to answer your question, for parts "a" and "b" there is intentionally no velocity vector at all since free body diagrams only show force vectors. In both parts "a" and "b", the friction force is directed up the ramp, and so the arrow representing these frictions forces point in the same direction. I suppose there is a case for making separate diagrams, but the distinction is only this: the kinetic force vector of part "b" should be shorter than the static friction force vector of part "a" since kinetic friction is always weaker than static friction.

Hi Ichbishop, thanks for the question. I suppose there's a bit of an oversight here, although it isn't a critical one since in the end we're looking for a ratio of masses, which we know must be positive. In the substitution shown in green, it would be slightly better to write $\dfrac{2V_2^{'2}}{v_1'} = \dfrac{-v_2'}{v_1'}$ in which case $\dfrac{v_2'}{v_1'} = \dfrac{-1}{2}$. Then the substitution into $\dfrac{m_1}{m_2} = \dfrac{-v_2'}{v_1'}$ would lead to the negatives cancelling, resulting in the expected positive final answer. Thanks for noticing that detail.

Why don't you use cosine theta in the work formula at the very end of the problem if W=Fdcos(theta)? instead of just d (just displacement) in incline problems like this?

Thanks for the question. I don't mean at all to contradict $W = Fd \cos(\theta)$ (which can also be written $W = F \cos(\theta) (d)$). That formula means that you multiply the force by the displacement, then multiply by cosine of the angle between them. In this particular solution, the force and displacement are actually parallel. The force I'm referring to is the force applied by some person or machine to make the car move up the ramp at constant speed. This applied force is directed along the ramp, so it's parallel to the length of the ramp. This means $W = F d \cos(0)$. Since $\cos(0) = 1$, it's common to not bother writing it at all, and to just write $W = F d$ for the special case where $\theta = 0$, which is the case where the force and displacement are parallel, as they are in this problem.

The applied force, according to the diagram, is $F_a = mg \sin(\theta)$, where theta here is referring to the angle of the incline. That's confusing, since the "theta" in the formula you wrote is referring to something different: namely the angle between the applied force and the displacement. If I were to write things out completely, using numbers in place of the angles, it would be $W = mg \sin(9.0^\circ) d \cos(0^\circ)$, where I added the $\cos(0^\circ)$ to try and be clear that that angle is the one in the $W = Fd\cos(\theta)$ formula, not the $9.0^\circ$ used to calculate the component of gravity along the incline.

Hi Icbishop, yes exactly. By "energy goes into heat" I mean to say "energy is transformed into heat energy". Rather than being gravitational potential energy, or kinetic energy, some of the energy is transformed into heat. This is the result of friction between particles in the ball when it gets squished upon colliding with the ground. The same could be said for particles in the ground internally "rubbing" and creating heat as the result of friction. Plus particles of the ground rub against particles of the ball. What I didn't mention is that some energy turns into sound energy, but this is typically very small compared to the amount turned into heat.

Thanks for the question. Only the change in height is important for questions involving gravitational potential energy, so the 8 degree incline is an extraneous detail. It's the vertical change in height that's important.

Is there anyway the solutions can be laid out onto one sheet so there isn't much scrolling? The calculations are nice and detailed but I lose my train of thought every time the scrolling happens. Makes it hard to follow the solutions -Alice Zhang

Hi tongmeng.zhang, yes I can see what you mean, especially on a 10min video. The video is a screen capture of my computer screen. I wonder if in the future I should change the orientation of the screen to be in "portrait" mode rather than "landscape"? It would sacrifice some width, but would keep more of the work on the screen at once. Unfortunately for the videos that are already made it would be difficult to change things. I could "zoom out", but then it would be too small to read. Thanks for the feedback though, since I'll try using "portrait" mode on future videos.

In the beginning when we solved for the Vx Initial and got 2.3077 why did we also use that in the magnitude calculation in the end? We solved for a final velocity of y, but we reused the velocity of x, could you please explain. Thanks.

Hi Icbishop, thanks for the question. Just to be clear, we're talking about part a), right? In part a) we solved for the resultant velocity, which is to say we solved for the vector sum of the x and y components. The answer for part a) is not the final velocity of y, but rather the final resultant velocity, which is normally just referred to as the velocity, but I'm including the word resultant to answer your question by explaining that the velocity is arrived at by using the pythagorean theorem to get it's magnitude by taking the square root of the sum of the squares of the x and y components of the velocity, and using trigonometry to get the direction.

Hi idan, I think it's a matter of personal opinion. Yes, using a graphing calculator is fine, but an "analytical solution" (ie. with algebra) can be useful too. The advantage of an algebra solution is that you can learn from an algebraic solution, and more generally characterize the behaviour of a system. Sorry to be so abstract, but we're talking about solutions in general, so it's difficult to be concrete. But consider the solution to problem 32 in chapter 4: $a = \dfrac{gm_B}{m_A + m_b}\textrm{ , } F_T = \dfrac{gm_Am_B}{m_A + m_B}$. That's an analytical solution, and it wouldn't be possible to use a graphing calculator since there are no numbers given in the problem. By looking at that analytical solution you can say things like "the acceleration decreases as $m_A$ increases", or "the tension force increases in proportion to both $m_A$ or $m_B$". Being able to create an analytical solution is important, so I guess my answer to your question is: yes, it's OK to use a graphing calculator, but make sure you could create an analytical solution if you had to, since that's your only option for problems with no numbers. It's like driving a race car: it's OK to drive it fast if you're a skilled driver, but not OK to drive fast if you're doing so just because you're unable to drive slow.

Thanks for the good question.
All the best,
Mr. Dychko

Hi Icbishop, thanks for the question. It's fine to choose the coordinate system you prefer (such as up as positive, which is conventional), and this will result in a correct calculation regardless of which choice you made, but you need to be aware of what the question is asking for. In this case the question is asking for the height of the cliff. Heights are always positive. They're a magnitude. This means that regardless of whether your calculation results in a positive or negative, you would give a positive as your final answer. The calculation gives you the displacement of the stone, but "displacement of the stone" is not what the question is asking for. The cliff height is the magnitude of the stone's displacement, which is to say the "absolute value", which are two fancy ways of saying "make it positive".

Thanks sanghoonwilliamk, you're totally right. $0.63 \textrm{ }\mu\textrm{F}$ is correct. While the working is correct in the video, I made a careless error turning the scientific notation into a decimal, and I put a note about this in the quick answer.

Thanks for the good question. Choosing whether the upward or downward direction is positive is called "choosing a coordinate system". Choosing a coordinate system is something you do for every solution, and the choice is a matter of personal preference. The only rule is that you're consistent, which means if you choose the downward acceleration to be positive, then it means downward velocity also must be positive, and the same goes for downward displacement. I choose down to be the positive direction since it meant there were no negative signs in my work, which I think looks just a little cleaner. The question is asking only for distances anyhow, which are positive, so I might as well make my calculations result in the final answer, rather than finding a negative downward displacement (which would have happened if I made the more conventional choice of upward being positive for the coordinate system) and then taking it's "magnitude" to get the final positive answer for height of the cliff.

Well I double checked and I do indeed have a part c on #42 chapter 2. It is the 7th Edition as well. The question also matches up with the solution provided.

## Giancoli 6th Edition, Chapter 7, Problem 35

By upasna.us1 on Sun, 01/08/2017 - 20:28I still don't understand why the initial velocity of the moving car is equal to the final velocity of both vehicles together? Could you explain this please?

## Giancoli 6th Edition, Chapter 2, Problem 44

By dmorochnick on Sat, 12/31/2016 - 07:40At about -2:05, shouldn't acceleration be negative: -9.8 yielding the solution d=4.3m instead of 2.1m?

Hi dmorochnick, thanks for the question. Earlier in the video I mentioned that the coordinate system was chosen so that

downis positive. While the convention is that if one says nothing about the coordinate system, the assumption is thatupis positive, is OK to explicitly change that. The choice is just personal preference. Where it's convenient, I sometimes prefer to make down positive to avoid substituting many negative values into the formula since I find so many negatives, and subtracting negatives, to be confusing. If you wish to haveupas positive, it's totally OK to do so. You would arrive at the same answer (not $4.3 \textrm{ m}$), since keep in mind that the displacement past the window would also have to be negative since it's downward (which in your coordinate system is the negative direction if up is positive).All the best,

Mr. Dychko

## Giancoli 6th Edition, Chapter 6, Problem 3

By nalakija on Thu, 12/08/2016 - 15:52Why are all the editions labeled as one??? Messing me up..

Hi nalakija, thanks for the comment. Could you please be more specific? I'm seeing editions labeled as "7th Edition", "6th Edition" and "Global Edition" in the menu bar. Where are you seeing them labelled as one?

Cheers,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 37

By jmarra_villanova on Wed, 12/07/2016 - 20:31Hi, how do you distinguish whether to use KEf + PEf = KEi + PEi on a spring question like this, rather than using Wext = DeltaPE in a question like #35?

Thank you

Hi jmarra_villanova,

Thanks for the question. The formula $KE_f + PE_f = KE_i + PE_i$ is a

law, which means it's always valid. This is an algebraic way of saying energy is conserved after a movement of some kind. As a law, it could also be used in #35 instead of $W_{ext} = \Delta PE$. Doing so would have looked like this: $0 + mgy_2 = 0 + mgy_1 + \dfrac{1}{2}k(y_2 - y_1)^2$, where the zero's are substituting for kinetic energy, of which there's none since in #35 the block isn't moving (or the movement is so slow as to be negligible), and $(y_2 - y_1)$ is the amount the spring was stretched. Using $W_{ext} = \Delta PE$ is just a convenience since it reduces a couple of algebra steps. When in doubt, use conservation of energy.Hope that helps,

Mr. Dychko

## Giancoli 7th "Global" Edition, Chapter 12, Problem 35

By maia.fehr.oth23 on Thu, 12/01/2016 - 21:44Why 331 and not 343?

## Giancoli 7th Edition, Chapter 10, Problem 40

By odelay.chewy on Fri, 11/25/2016 - 00:47I am struggling understanding the funky algebra used in this step. You kind of gloss over it. Could you show it step for step?

Hi odelay.chewy,

Thanks for the question. I'm away from my recording equipment at the moment, but I'll try to give some highlights here. We start with saying the total forces directed up equal the total forces directed down, then I make a substitution for each up force with the archimedes principle that the buoyant force is the weight of fluid displaced, which is $F = \rho V g$, and the down forces are the usual weight formula $F = mg$. Then, in green, the volume of the wood and lead is expressed in terms of their density and mass. This is in turn substituted (using red) for each volume factor in the up force terms. The specific gravity of wood replaces the ratio of the density of water to wood. Then comes a confusing line where the mass of the wood is shown moving to the left, but actually ends up on the right. The thing to notice with that line is that we're collecting like terms. We're placing the wood terms together on one side, and the lead terms on the other side. The mass of lead and mass of wood are factored out on their respective sides, leaving us with some bracket multiplied by a mass on each side. We're interested in knowing the mass of lead, so we divide both sides by the bracket by which the lead mass is multiplied. Then we plug and chug. Hopefully this video commentary is a bit helpful...

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 13, Problem 14

By saie.joshi on Thu, 11/24/2016 - 08:42Why do you subtract the values instead of adding them together? Don't both the water and the glass reduce volume?

never mind I got it!

## Giancoli 7th Edition, Chapter 8, Problem 22

By taylorreilley on Mon, 11/14/2016 - 08:32Why did you plug the Wf value in for b when you wrote Wi

Hi taylerreilley,

Thanks for the question. "initial" and "final" is a bit confusing in this questions since there are more than one time periods to consider, each with their own "initial" and "final" moment. For part a), the "final" moment is when the car has reached $55 \textrm{ km/h}$, making $\omega_f = 38.194 \textrm{ rad/s}$. However, for part b), the

initialmoment is when the car is at $55 \textrm{ km/h}$, so for part b) $\omega_i = 38.194 \textrm{ rad/s}$. The initial moment of part b) is the final moment of part a).I can see where you're coming from, and technically it would be more correct for me to put a subscript on the variables to assign them to parts of the question, such as for the angular speed for part a) written as $\omega_{iA}$ and $\omega_{fA}$, and then say $\omega_{fA} = \omega_{iB}$, and so on, but I think that also creates it's own source of confusion by presenting so many more subscripts.

All the best,

Mr. Dyckho

## Giancoli 7th Edition, Chapter 7, Problem 41

By sheumangutman on Fri, 11/11/2016 - 16:12Why is this collision deemed "inelastic" in terms of the treatment of kinetic energy conservation. Seems based on the diagram, the objects are not sticking together, which would suggest its inelastic.

Hi sheumangutman,

Thanks for the question. The term "inelastic" means there is less total kinetic energy after a collision compared with before a collision. Since that's the case here, we can call it inelastic. It is not "completely inelastic", that's true. A completely inelastic collision involves the two particles sticking together, and this type of collision results in the maximum loss in kinetic energy that's possible while still conserving momentum. Seeing two particles bounce apart means either the collision is "inelastic" (but not "completely inelastic"), or maybe the collision is "perfectly elastic", in which case the total kinetic energy after collision is the same as it was before. In this case, the collision is a regular inelastic collision since some kinetic energy was lost, but not the maximum loss that would have otherwise been possible had they stuck together.

Best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 9

By sheumangutman on Thu, 11/10/2016 - 08:28Could you also approach the problem by first solving for final velocity by using V=a*t, and then plugging this velocity into v^2=vo^2+2AD to solve for displacement?

Thanks.

Hi sheumangutman,

Yes you can! Substituting $v=at$ info $v^2 = v_o^2 + 2aD$ gives $a^2t^2 = v_o^2 + 2aD$, which, with $v_o = 0$ rearranges to $D = \dfrac{1}{2}at^2$ after you divide both sides by $2a$. $D = \dfrac{1}{2}at^2$ is the formula I used in the video. Good work!

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 52

By schamorg on Wed, 11/09/2016 - 07:09Thank you so much. This was very helpful actually.

Super! I'm glad, and thanks for the feedback. :)

## Giancoli 7th Edition, Chapter 3, Problem 37

By kmoons25 on Mon, 11/07/2016 - 20:00why do you do these things?

Hi kmoons25, Well, I know physics can be confusing. When you have a more specific question, just let me know.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 53

By taylorreilley on Sun, 11/06/2016 - 15:46Why is there still potential energy at the end?

Hi taylorreilley,

Thanks for the question. One thing to understand about potential energy is that you can never know how much there is. :) There's no such thing as an "absolute" potential energy. It's possible only to tell what is the

changein potential energy. Sometimes we speak as though there is an absolute potential energy, such as "the book is 3m above the ground, so it has $mg(3 \textrm{ m})$ of potential energy, but don't be fooled by that as it's just a concise (maybe lazy?) way of speaking, when in fact it means thedifferencein potential energy of the bookwith respect tothe ground is $mg(3 \textrm{ m})$. With that in mind, the skier has achangein potential energy resulting from their drop in altitude of $230 \textrm{ m}$, and thischangein potential energy is turned into other forms of energy, namely kinetic energy and heat. Subtracting away the kinetic energy gained from the loss of potential energy, we're left with how much of that potential energy loss was turned into heat.To put it in one sentence, the "potential energy" term you see in the final calculation is not the "potential energy at the end", but rather, it is the

changein potential energy when the skier descends.All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 9, Problem 28

By leopinzon18 on Sat, 11/05/2016 - 10:02what formulas are being used

Hi leopinzon18, for formulas being used are $\tau = Fl_{\bot}$ which says torque equals force times the perpendicular component of the lever arm, and the other formula is that $\Sigma \tau = 0$ for static equilibrium, which is also to say $\Sigma \tau_{ccw} = \Sigma \tau_{cw}$ meaning the counter-clockwise torques equal the clockwise torques. If you have a more specific question, just let me know and I'll see if I can answer it.

Best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 52

By schamorg on Thu, 11/03/2016 - 19:43Hello!

I am still confused as to why the diagrams for a) and b) are the same. I don't understand how in a) the box has static friction, but b) has kinetic friction and their diagrams are the same. Shouldn't b) have a velocity vector pointing downward?

Thank you so much for your time!

Hi schamorg, thanks for asking this question. Understanding free body diagrams is really important, so it's good your digging into the topic. A free body diagram does not include a velocity vector. I sketched one in for part "c", but that was just to emphasize which direction the box is moving in order to explain why the force arrows are pointing the way they are. Notice that the velocity vector does not begin on the center of the box, since I didn't want to make is seem like another force vector. Force vectors are always written with their beginning a the center of mass of the object (usually the geometric center... just ignore this point if it's confusing). Technically, velocity vectors do not belong on free body diagrams. OK, so to answer your question, for parts "a" and "b" there is intentionally no velocity vector at all since free body diagrams only show force vectors. In both parts "a" and "b", the friction force is directed up the ramp, and so the arrow representing these frictions forces point in the same direction. I suppose there is a case for making separate diagrams, but the distinction is only this: the kinetic force vector of part "b" should be shorter than the static friction force vector of part "a" since kinetic friction is always weaker than static friction.

Hope this helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 7, Problem 10

By lcbishop on Wed, 11/02/2016 - 11:51Why is the negative sign from the v2 prime early not considered in the final calculation. Thanks.

Hi Ichbishop, thanks for the question. I suppose there's a bit of an oversight here, although it isn't a critical one since in the end we're looking for a ratio of masses, which we know must be positive. In the substitution shown in green, it would be slightly better to write $\dfrac{2V_2^{'2}}{v_1'} = \dfrac{-v_2'}{v_1'}$ in which case $\dfrac{v_2'}{v_1'} = \dfrac{-1}{2}$. Then the substitution into $\dfrac{m_1}{m_2} = \dfrac{-v_2'}{v_1'}$ would lead to the negatives cancelling, resulting in the expected positive final answer. Thanks for noticing that detail.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 5

By lyonsca1 on Fri, 10/28/2016 - 08:23Why don't you use cosine theta in the work formula at the very end of the problem if W=Fdcos(theta)? instead of just d (just displacement) in incline problems like this?

Hi lyonsca1,

Thanks for the question. I don't mean at all to contradict $W = Fd \cos(\theta)$ (which can also be written $W = F \cos(\theta) (d)$). That formula means that you multiply the force by the displacement, then multiply by cosine of the angle between them. In this particular solution, the force and displacement are actually parallel. The force I'm referring to is the force applied by some person or machine to make the car move up the ramp at constant speed. This applied force is directed along the ramp, so it's parallel to the length of the ramp. This means $W = F d \cos(0)$. Since $\cos(0) = 1$, it's common to not bother writing it at all, and to just write $W = F d$ for the special case where $\theta = 0$, which is the case where the force and displacement are parallel, as they are in this problem.

The applied force, according to the diagram, is $F_a = mg \sin(\theta)$, where theta here is referring to the angle of the incline. That's confusing, since the "theta" in the formula you wrote is referring to something different: namely the angle between the applied force and the displacement. If I were to write things out completely, using numbers in place of the angles, it would be $W = mg \sin(9.0^\circ) d \cos(0^\circ)$, where I added the $\cos(0^\circ)$ to try and be clear that that angle is the one in the $W = Fd\cos(\theta)$ formula, not the $9.0^\circ$ used to calculate the component of gravity along the incline.

Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 52

By lcbishop on Mon, 10/24/2016 - 11:35Can you give a brief explanation of what you mean by "energy goes into heat"? Is it transformed into heat? Thanks.

Hi Icbishop, yes exactly. By "energy goes into heat" I mean to say "energy is transformed into heat energy". Rather than being gravitational potential energy, or kinetic energy, some of the energy is transformed into heat. This is the result of friction between particles in the ball when it gets squished upon colliding with the ground. The same could be said for particles in the ground internally "rubbing" and creating heat as the result of friction. Plus particles of the ground rub against particles of the ball. What I didn't mention is that some energy turns into sound energy, but this is typically very small compared to the amount turned into heat.

Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 31

By jessicasingh001 on Sun, 10/23/2016 - 18:46Does anyone know why the 8 degree incline is ignored?

Hi jessicasingh001,

Thanks for the question. Only the change in height is important for questions involving gravitational potential energy, so the 8 degree incline is an extraneous detail. It's the vertical change in height that's important.

Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 7, Problem 28

By gillander97 on Sun, 10/23/2016 - 15:26no sound ;-(

Hi gellander97, thanks for the comment. The sound plays fine on this end, so just let me know if that happens again.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 66

By tongmeng.zhang on Sat, 10/15/2016 - 11:43Is there anyway the solutions can be laid out onto one sheet so there isn't much scrolling? The calculations are nice and detailed but I lose my train of thought every time the scrolling happens. Makes it hard to follow the solutions -Alice Zhang

Hi tongmeng.zhang, yes I can see what you mean, especially on a 10min video. The video is a screen capture of my computer screen. I wonder if in the future I should change the orientation of the screen to be in "portrait" mode rather than "landscape"? It would sacrifice some width, but would keep more of the work on the screen at once. Unfortunately for the videos that are already made it would be difficult to change things. I could "zoom out", but then it would be too small to read. Thanks for the feedback though, since I'll try using "portrait" mode on future videos.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 3, Problem 33

By lcbishop on Thu, 10/13/2016 - 21:31In the beginning when we solved for the Vx Initial and got 2.3077 why did we also use that in the magnitude calculation in the end? We solved for a final velocity of y, but we reused the velocity of x, could you please explain. Thanks.

Hi Icbishop, thanks for the question. Just to be clear, we're talking about part a), right? In part a) we solved for the

resultant velocity, which is to say we solved for the vector sum of thexandycomponents. The answer for part a) is not the final velocity of y, but rather the finalresultant velocity, which is normally just referred to as thevelocity, but I'm including the wordresultantto answer your question by explaining that the velocity is arrived at by using the pythagorean theorem to get it's magnitude by taking the square root of the sum of the squares of the x and y components of the velocity, and using trigonometry to get the direction.Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 2, Problem 49

By idan on Thu, 10/13/2016 - 17:13Do you think it's acceptable to use a graphing calculator to find the zeros of these functions? We're studying physics, not math.

Hi idan, I think it's a matter of personal opinion. Yes, using a graphing calculator is fine, but an "analytical solution" (ie. with algebra) can be useful too. The advantage of an algebra solution is that you can learn from an algebraic solution, and more generally characterize the behaviour of a system. Sorry to be so abstract, but we're talking about solutions in general, so it's difficult to be concrete. But consider the solution to problem 32 in chapter 4: $a = \dfrac{gm_B}{m_A + m_b}\textrm{ , } F_T = \dfrac{gm_Am_B}{m_A + m_B}$. That's an analytical solution, and it wouldn't be possible to use a graphing calculator since there are no numbers given in the problem. By looking at that analytical solution you can say things like "the acceleration decreases as $m_A$ increases", or "the tension force increases in proportion to both $m_A$ or $m_B$". Being able to create an analytical solution is important, so I guess my answer to your question is: yes, it's OK to use a graphing calculator, but make sure you

couldcreate an analytical solution if you had to, since that's your only option for problems with no numbers. It's like driving a race car: it's OK to drive it fast if you're a skilled driver, but not OK to drive fast if you're doing so just because you're unable to drive slow.Thanks for the good question.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 2, Problem 39

By lcbishop on Sat, 10/08/2016 - 17:17If for instance we considered down not to be positive and ended up with a negative. Does this essentially make the answer wrong?

Hi Icbishop, thanks for the question. It's fine to choose the coordinate system you prefer (such as up as positive, which is conventional), and this will result in a correct calculation regardless of which choice you made, but you need to be aware of what the question is asking for. In this case the question is asking for the

heightof the cliff. Heights are always positive. They're amagnitude. This means that regardless of whether your calculation results in a positive or negative, you would give a positive as your final answer. The calculation gives you the displacement of the stone, but "displacement of the stone" is not what the question is asking for. The cliff height is themagnitudeof the stone's displacement, which is to say the "absolute value", which are two fancy ways of saying "make it positive".Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 18, Problem 3

By daniel.weiss1 on Fri, 09/30/2016 - 11:02Great video, small note: I think you mistakenly have coulombs in your answer above the video instead of amps.

Thanks daniel.weiss1! I just fixed it.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 2, Problem 41

By sheumangutman on Tue, 09/27/2016 - 06:57Why cant the formula v1=vo+At be used for this problem assuming v1=0?

## Giancoli 7th Edition, Chapter 3, Problem 42

By julia.wolfe on Mon, 09/26/2016 - 12:54At 1:45 why is the x component 4.24 + 1.7 shown as equaling 2.124? Shouldn't it be 5.94 since the stair climb and boat direction are the same?

Oops nevermind! I had copied some numbers wrong!

## Giancoli 7th Edition, Chapter 17, Problem 39

By sanghoonwilliamk on Sat, 09/24/2016 - 21:45i think this should be .63 microfarads

Thanks sanghoonwilliamk, you're totally right. $0.63 \textrm{ }\mu\textrm{F}$ is correct. While the working is correct in the video, I made a careless error turning the scientific notation into a decimal, and I put a note about this in the quick answer.

Thanks for the sharp eye!

Mr. Dychko

Okay gotcha~ thanks for the clarification.

## Giancoli 7th Edition, Chapter 3, Problem 18

By sheumangutman on Sat, 09/24/2016 - 19:00Can you please explain why the acceleration in this question is assumed to be positive while it was negative in question #17? Thanks.

Hi sheumangutman,

Thanks for the good question. Choosing whether the upward or downward direction is positive is called "choosing a coordinate system". Choosing a coordinate system is something you do for every solution, and the choice is a matter of personal preference. The only rule is that you're consistent, which means if you choose the downward acceleration to be positive, then it means downward velocity also must be positive, and the same goes for downward displacement. I choose down to be the positive direction since it meant there were no negative signs in my work, which I think looks just a little cleaner. The question is asking only for distances anyhow, which are positive, so I might as well make my calculations result in the final answer, rather than finding a negative downward displacement (which would have happened if I made the more conventional choice of upward being positive for the coordinate system) and then taking it's "magnitude" to get the final positive answer for height of the cliff.

Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 2, Problem 42

By lcbishop on Sat, 09/24/2016 - 12:13For part c) What specific factors make this an estimate?

Hi Icbishop, did you post this question on the wrong video? I don't see a part c) for this problem.

Cheers,

Mr. Dychko

Well I double checked and I do indeed have a part c on #42 chapter 2. It is the 7th Edition as well. The question also matches up with the solution provided.