Hi idan, you are making a valid point. You have a sharp eye! Mathematically, it would be more correct to factor and then find the roots of the resulting equation, thereby discovering the answer $t=0$. However, since this question is asking for the "time in the air", $t=0$, as you know, is an extraneous solution, so the effort of finding the solution $t=0$ and then discarding it as extraneous isn't worth the effort. The technique in the video is perfectly fine for a physics class, since math is just a tool for finding solutions to the physical problem, which in this case is "how much time is the ball in the air", not "at what times is the ball at height zero". For the latter, one must follow the technique you suggest by factoring since $t=0$ would be a non-trivial solution, but as it is, $t=0$ is not a solution for determining the "time in the air" problem. In a math class there's no question that to answer "solve this equation", one must factor and include $t=0$ as a solution.

Hi idan, you're quite right that the kinematics equations contain displacement, rather than distance. Often times, such as in this question, the two quantities have the same magnitude. The distance and the magnitude of the displacement are the same, in other words, but your comment and awareness that d is in fact always displacement is definitely important in questions for which that's not the case. Technically it's a mistake for me to refer to the factor d as a distance, but I fear that I probably do it quite often, making the distinction mostly just where it's necessary to do so. Hopefully this isn't too confusing!

Hello student from thesoutportschool, thank you for spotting the error. At 3:23 I mention correctly that the angular velocity is $3.333 \times 10^{-4}$, whereas at 3:46 I incorrectly wrote $3.333 \times 10^{-11}$. My calculations were nevertheless done using $3.333 \times 10^{-4}$, so the results are correct. I'll put a note above the video to use $3.333 \times 10^{-4}$ for the angular velocity.

Hi margolinw, I think you're referring to the algebra step at 3:36, right? Multiplying by $\dfrac{r^2}{10^{\beta/10}}$ is a way to isolate the unknown $r$ by itself on one side of the equation. Multiplying by $r^2$ cancels it on the left side, whereas dividing by $10^{\beta/10}$ cancels that term on the right side, and this results in $r^2$ by itself on the right side. Mission accomplished! Then I switched the sides around and took the square root of $r^2$.

How are you supposed to know that the comb gained (rather than lost) electrons? I know that the chapter mentions that plastic items develop negative charges when rubbed, but I thought the problem would specify that the charge acquired was -3 microcoulombs if the comb had gained (vs. lost) electrons?
Thanks,
Julia

I think the question would be worded better if it said $-3.0 \textrm{ } \mu \textrm{C}$. I don't think you're meant to know that plastic gains electrons, although in this case, without the question specifying a negative charge, knowing the properties of plastic is the only way to know. I would presume that in any real test question that the negative sign would be given.

Thanks for the question. Yes, things are all good here. The calculator display shows /4/pi^2, which is another way of saying what you might have been expecting: /(4pi^2).

Hi tuh20232, thanks for the question. P = the power consumed, always. On the packaging for a bulb, the manufacturer will give the answer to the power calculation, assuming a particular voltage. For a bulb sold in the US, the manufacturer will assume 120 V in the calculation for the bulb power, where for a bulb sold in the EU, the manufacturer will assume 240 V.

P = the power consumed, whereas the P = ## rating for a bulb depends on where the bulb is sold.

Thanks for the question. This diagram is a bit tricky since some resistors are in series, whereas others are in parallel. We break it down step by step. If current reaching a node has a choice between multiple paths, then the resistors along the two (or more) paths are in parallel. If current reaching a node has only one option, and that is to proceed through another resistor, then that next resistor is in series. Does that help?

In the video, you wrote 1.29 * 10^5 Pa, but in the main answer section, you wrote 1.20 * 10^5 Pa. Merely for clarity, can you correct the second answer?

Hi m_iqbal, thanks for the question. The negative sign for velocity indicates direction only. It says nothing about whether the object is speeding up or slowing down. A negative velocity means only that the object is moving in the negative direction, which is typically to the left, although a person solving a problem is free to redefine the coordinate system by saying "I feel like making the negative direction to the right". It's a bit unusual to say "right is negative", but you do need to be prepared for this idea since occasionally I do say "let down be the positive direction", rather than the traditional "up is positive", when doing so simplifies the mathematics by avoiding so many negative signs in a formula.

A change in speed is given by acceleration, so your next question might be: does a negative acceleration mean the object is speeding up or slowing down. The answer here depends on whether the object currently has a negative velocity, or not. If the acceleration has the same sign as the velocity, it's speeding up. An acceleration with an opposite sign to the velocity means the object is slowing down. So, running through the list of possibilities: +a & +v => speeding up, -a & +v => slowing down, +a & -v => slowing down, -a & -v => speeding up. The last scenario is probably the one you'll want to take note of, where a negative acceleration means the object is speeding up when it already has negative velocity.

Thank you very much jsh2672 for noticing that! I've corrected the quick answer to show the result after multiplying by 100, and put a note about the error in the video.

Hi! Would you be able to explain something to me? I don't understand why solving for that x component gives the average horizontal speed. I thought that x component is only applicable to the instantaneous initial velocity?

Thanks a lot for the question, and I'm sorry for taking so long to get back. I hope you're still working on this unit....

Sure, it's true that the x-component of the velocity is that of the instantaneous initial velocity, as you say. However, the x-component of the velocity never changes. There is no horizontal acceleration. Since the x-component of the velocity never changes, this means that whatever value it has initially will also be the value it has at any other time, and so in this circumstance it's initial value is also it's average. The average of a value that's constant is whatever that value is, at any time.

In part a) I'm confused as to why By is negative.
sin 56 = By/(-26.5) will result in a negative - I get that - but in question number 9, which uses the same vectors, the solution to By was positive, so why aren't they the same?

Is it because in the equation we're asked to solve for, we have to subtract vector B? Then why isn't Bx negative?

In question 13 the vector B is the negative of what it was in problem #9, which makes it point in the opposite direction. When I see a vector being subtracted, as vector B is in this case, I like to think of it as 'adding it's negative', which mean apply the usual rules for adding vectors, but flip around the one that's being subtracted. Instead of up and to the left in #9, it is now down and to the right here in #13. Since it's pointing downish, that makes the y-component, $B_y$, negative.

In both cases, the calculation for $B_y$ was $26.5 \sin(56^\circ)$ but here in #13 the result gets a negative in front if it despite the calculator saying 'positive'. This has to do with our understanding of the physical situation, in that the vector is pointing down, and so with that understanding we just put the negative there, regardless of what the calculator says. This might be going on a tangent, but just for arguments sake, if you wanted to always have the calculator say the correct sign, you would have to enter angles in standard position which is measured counter clockwise starting from the positive x-axis. In problem 13 the standard position angle of vector B is $360 - 56 = 304^\circ$, in which case $26.5 \times \sin(304^\circ) = -21.97...$ with the negative sign, rather than $26.5 \times \sin(124^\circ) = +21.97...$ (where $124^\circ$ is the standard position angle in #9). I don't recommend changing your angles to standard position, but I just mention it so that you're more comfortable placing negative signs where they belong 'by decrie', secure in knowing that there is a way to make the mathematics consistent with the physical reality if you felt like it. You're not doing anything wrong, in other words, by overriding the sign of what the calculator says and choosing the sign that fits the physical situation.

I guess I missed this in chapter 2, but why do we square the magnitude of the velocity when using the kinematic equation (when trying to solve for the acceleration, blue writing, 2 minutes 3 seconds in)? It's a mistake I keep making but I can't figure it out why.

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.

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

By idan on Sun, 02/26/2017 - 14:42By dividing by "t" instead of factoring aren't we losing a physically meaningful answer t=0 (the time the ball was hit)?

Hi idan, you are making a valid point. You have a sharp eye! Mathematically, it would be more correct to factor and then find the roots of the resulting equation, thereby discovering the answer $t=0$. However, since this question is asking for the "time in the air", $t=0$, as you know, is an extraneous solution, so the effort of finding the solution $t=0$ and then discarding it as extraneous isn't worth the effort. The technique in the video is perfectly fine for a physics class, since math is just a tool for finding solutions to the physical problem, which in this case is "how much time is the ball in the air", not "at what times is the ball at height zero". For the latter, one must follow the technique you suggest by factoring since $t=0$ would be a non-trivial solution, but as it is, $t=0$ is not a solution for determining the "time in the air" problem. In a math class there's no question that to answer "solve this equation", one must factor and include $t=0$ as a solution.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 2, Problem 26

By idan on Sun, 02/26/2017 - 12:28I noticed you called "d" distance, but aren't all the kinematic equations actually only referring to displacement?

Hi idan, you're quite right that the kinematics equations contain

displacement, rather than distance. Often times, such as in this question, the two quantities have the same magnitude. The distance and the magnitude of the displacement are the same, in other words, but your comment and awareness thatdis in fact always displacement is definitely important in questions for which that's not the case. Technically it's a mistake for me to refer to the factordas a distance, but I fear that I probably do it quite often, making the distinction mostly just where it's necessary to do so. Hopefully this isn't too confusing!All the best,

Mr. Dychko

## Giancoli 6th Edition, Chapter 8, Problem 64

By thesouthportschool on Sat, 02/18/2017 - 15:37Is the answer not 453.6926 J or 5*10^2 J, I copied the exact calculations from your final answer of part A (3:49) and did not get 5*10^16 J.

And I also you changed the angular velocity from 3.333*10^-4 to 3.333*10^-11 when doing the calculations to work out the KE

Hello student from thesoutportschool, thank you for spotting the error. At 3:23 I mention correctly that the angular velocity is $3.333 \times 10^{-4}$, whereas at 3:46 I incorrectly wrote $3.333 \times 10^{-11}$. My calculations were nevertheless done using $3.333 \times 10^{-4}$, so the results are correct. I'll put a note above the video to use $3.333 \times 10^{-4}$ for the angular velocity.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 12, Problem 15

By margolinw on Fri, 02/17/2017 - 08:04What is your rationale for deciding to multiply by (r^2/10^beta/10) in part B?

Hi margolinw, I think you're referring to the algebra step at 3:36, right? Multiplying by $\dfrac{r^2}{10^{\beta/10}}$ is a way to isolate the unknown $r$ by itself on one side of the equation. Multiplying by $r^2$ cancels it on the left side, whereas dividing by $10^{\beta/10}$ cancels that term on the right side, and this results in $r^2$ by itself on the right side. Mission accomplished! Then I switched the sides around and took the square root of $r^2$.

Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 16, Problem 5

By julia.wolfe on Sun, 02/12/2017 - 12:41How are you supposed to know that the comb gained (rather than lost) electrons? I know that the chapter mentions that plastic items develop negative charges when rubbed, but I thought the problem would specify that the charge acquired was -3 microcoulombs if the comb had gained (vs. lost) electrons?

Thanks,

Julia

Hi julia.wolfe,

I think the question would be worded better if it said $-3.0 \textrm{ } \mu \textrm{C}$. I don't think you're meant to know that plastic gains electrons, although in this case, without the question specifying a negative charge, knowing the properties of plastic is the only way to know. I would presume that in any real test question that the negative sign would be given.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 11, Problem 28

By donnarrnc on Fri, 02/10/2017 - 18:10Is this calculation correct? It shows 4/pi^2 on the calculator display.

Hi donnarrnc,

Thanks for the question. Yes, things are all good here. The calculator display shows /4/pi^2, which is another way of saying what you might have been expecting: /(4pi^2).

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 18, Problem 36

By tuh20232 on Thu, 02/09/2017 - 15:50I got confused with the P= 75 w and the P = the power consumed would you please explain the difference? Thank you

Hi tuh20232, thanks for the question. P = the power consumed, always. On the packaging for a bulb, the manufacturer will give the answer to the power calculation, assuming a particular voltage. For a bulb sold in the US, the manufacturer will assume 120 V in the calculation for the bulb power, where for a bulb sold in the EU, the manufacturer will assume 240 V.

P = the power consumed, whereas the P = ## rating for a bulb depends on where the bulb is sold.

Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 19, Problem 18

By saie.joshi on Sun, 02/05/2017 - 11:21do the nodes initially not mean anything then? I thought the nodes would make the resistors not in parallel or series.

Hi saie.joshi,

Thanks for the question. This diagram is a bit tricky since some resistors are in series, whereas others are in parallel. We break it down step by step. If current reaching a node has a choice between multiple paths, then the resistors along the two (or more) paths are in parallel. If current reaching a node has only one option, and that is to proceed through another resistor, then that next resistor is in series. Does that help?

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 10, Problem 22

By rdattafl on Thu, 02/02/2017 - 11:00In the video, you wrote 1.29 * 10^5 Pa, but in the main answer section, you wrote 1.20 * 10^5 Pa. Merely for clarity, can you correct the second answer?

Thank you very much for spotting that rdattafl. I've updated the quick answer.

## Giancoli 7th Edition, Chapter 2, Problem 30

By m_iqbal on Tue, 01/31/2017 - 19:20Hi,

I have a question that if the object has constant negative velocity, will the object speed up or slow down?

Hi m_iqbal, thanks for the question. The negative sign for velocity indicates

directiononly. It says nothing about whether the object is speeding up or slowing down. A negative velocity means only that the object is moving in the negative direction, which is typically to the left, although a person solving a problem is free to redefine the coordinate system by saying "I feel like making the negative direction to the right". It's a bit unusual to say "right is negative", but you do need to be prepared for this idea since occasionally I do say "let down be the positive direction", rather than the traditional "up is positive", when doing so simplifies the mathematics by avoiding so many negative signs in a formula.A change in speed is given by acceleration, so your next question might be: does a negative acceleration mean the object is speeding up or slowing down. The answer here depends on whether the object currently has a negative

velocity, or not. If the acceleration has the same sign as the velocity, it's speeding up. An acceleration with an opposite sign to the velocity means the object is slowing down. So, running through the list of possibilities: +a & +v => speeding up, -a & +v => slowing down, +a & -v => slowing down, -a & -v => speeding up. The last scenario is probably the one you'll want to take note of, where a negative acceleration means the object is speeding up when it already has negative velocity.All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 9, Problem 48

By jsh2672 on Tue, 01/31/2017 - 15:56Why are you multiplying by 1/100? Shouldn't you be multiplying by 100 to get the percentage?

Thank you very much jsh2672 for noticing that! I've corrected the quick answer to show the result after multiplying by 100, and put a note about the error in the video.

Best wishes,

Mr. Dychko

## Giancoli 7th Edition, Chapter 3, Problem 25

By sophiaswimgirl on Tue, 01/24/2017 - 23:29Hi! Would you be able to explain something to me? I don't understand why solving for that x component gives the average horizontal speed. I thought that x component is only applicable to the instantaneous initial velocity?

Hi sophiaswimgirl,

Thanks a lot for the question, and I'm sorry for taking so long to get back. I hope you're still working on this unit....

Sure, it's true that the x-component of the velocity is that of the instantaneous initial velocity, as you say. However, the x-component of the velocity never changes. There is no horizontal acceleration. Since the x-component of the velocity never changes, this means that whatever value it has initially will also be the value it has at any other time, and so in this circumstance it's initial value is also it's average. The average of a value that's constant is whatever that value is, at any time.

Hope this helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 3, Problem 13

By sophiaswimgirl on Fri, 01/20/2017 - 17:16In part a) I'm confused as to why By is negative.

sin 56 = By/(-26.5) will result in a negative - I get that - but in question number 9, which uses the same vectors, the solution to By was positive, so why aren't they the same?

Is it because in the equation we're asked to solve for, we have to subtract vector B? Then why isn't Bx negative?

Thank you.

Hi sophiaswimgirl,

In question 13 the vector

Bis the negative of what it was in problem #9, which makes it point in the opposite direction. When I see a vector being subtracted, as vectorBis in this case, I like to think of it as 'adding it's negative', which mean apply the usual rules for adding vectors, but flip around the one that's being subtracted. Instead of up and to the left in #9, it is now down and to the right here in #13. Since it's pointing downish, that makes the y-component, $B_y$, negative.In both cases, the calculation for $B_y$ was $26.5 \sin(56^\circ)$ but here in #13 the result gets a negative in front if it despite the calculator saying 'positive'. This has to do with our understanding of the physical situation, in that the vector is pointing down, and so with that understanding we just put the negative there, regardless of what the calculator says. This might be going on a tangent, but just for arguments sake, if you wanted to always have the calculator say the correct sign, you would have to enter angles in standard position which is measured counter clockwise starting from the positive x-axis. In problem 13 the standard position angle of vector

Bis $360 - 56 = 304^\circ$, in which case $26.5 \times \sin(304^\circ) = -21.97...$ with the negative sign, rather than $26.5 \times \sin(124^\circ) = +21.97...$ (where $124^\circ$ is the standard position angle in #9). I don't recommend changing your angles to standard position, but I just mention it so that you're more comfortable placing negative signs where they belong 'by decrie', secure in knowing that there is a way to make the mathematics consistent with the physical reality if you felt like it. You're not doing anything wrong, in other words, by overriding the sign of what the calculator says and choosing the sign that fits the physical situation.Maybe that was a bit long, but I hope it helps.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 5

By sophiaswimgirl on Thu, 01/19/2017 - 00:23I guess I missed this in chapter 2, but why do we square the magnitude of the velocity when using the kinematic equation (when trying to solve for the acceleration, blue writing, 2 minutes 3 seconds in)? It's a mistake I keep making but I can't figure it out why.

Thanks so much for your help.

Wait, I figured it out. It was staring me right in the face - I had the wrong formula. Sorry, thanks!

## Giancoli 6th Edition, Chapter 14, Problem 25

By sulaiman_gos on Tue, 01/10/2017 - 12:21hi .. I have a question here .. why you keep solving with ΔT as it's ( Ti - Tf ) but it is actually ΔT = ( Tf - Ti ) !? please I need explanation!!

## Giancoli 7th Edition, Chapter 17, Problem 2

By a_murayyan on Tue, 01/10/2017 - 06:51in the answer box, don't you mean 2.72 X 10^-17 J?

## 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