Comments

Giancoli 7th Edition, Chapter 9, Problem 48

By jsh2672 on Tue, 01/31/2017 - 15:56

Why are you multiplying by 1/100? Shouldn't you be multiplying by 100 to get the percentage?

By Mr. Dychko on Sun, 02/05/2017 - 12:43

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 - 23:29

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?

By Mr. Dychko on Sun, 02/05/2017 - 12:50

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 - 17:16

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?

Thank you.

By Mr. Dychko on Sun, 02/05/2017 - 13:19

Hi sophiaswimgirl,

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.

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 - 00:23

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.

Thanks so much for your help.

By sophiaswimgirl on Thu, 01/19/2017 - 00:24

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 - 12:21

hi .. 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 - 06:51

in 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 - 20:28

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?

Giancoli 6th Edition, Chapter 2, Problem 44

By dmorochnick on Sat, 12/31/2016 - 07:40

At about -2:05, shouldn't acceleration be negative: -9.8 yielding the solution d=4.3m instead of 2.1m?

By Mr. Dychko on Sun, 01/01/2017 - 01:15

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).

All the best,
Mr. Dychko

Giancoli 6th Edition, Chapter 6, Problem 3

By nalakija on Thu, 12/08/2016 - 15:52

Why are all the editions labeled as one??? Messing me up..

By Mr. Dychko on Thu, 12/08/2016 - 23:53

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 - 20:31

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

By Mr. Dychko on Thu, 12/08/2016 - 23:51

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 - 21:44

Why 331 and not 343?

Giancoli 7th Edition, Chapter 10, Problem 40

By odelay.chewy on Fri, 11/25/2016 - 00:47

I am struggling understanding the funky algebra used in this step. You kind of gloss over it. Could you show it step for step?

By Mr. Dychko on Sat, 11/26/2016 - 11:39

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 - 08:42

Why do you subtract the values instead of adding them together? Don't both the water and the glass reduce volume?

By saie.joshi on Thu, 11/24/2016 - 08:43

never mind I got it!

Giancoli 7th Edition, Chapter 8, Problem 22

By taylorreilley on Mon, 11/14/2016 - 08:32

Why did you plug the Wf value in for b when you wrote Wi

By Mr. Dychko on Wed, 11/23/2016 - 00:32

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 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.

All the best,
Mr. Dyckho

Giancoli 7th Edition, Chapter 7, Problem 41

By sheumangutman on Fri, 11/11/2016 - 16:12

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.

By Mr. Dychko on Sat, 11/12/2016 - 13:16

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 - 08:28

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?

Thanks.

By Mr. Dychko on Sat, 11/12/2016 - 13:03

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 - 07:09

Thank you so much. This was very helpful actually.

By Mr. Dychko on Wed, 11/09/2016 - 07:11

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

Giancoli 7th Edition, Chapter 3, Problem 37

By kmoons25 on Mon, 11/07/2016 - 20:00

why do you do these things?

By Mr. Dychko on Tue, 11/08/2016 - 00:54

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 - 15:46

Why is there still potential energy at the end?

By Mr. Dychko on Tue, 11/08/2016 - 00:53

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 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.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 9, Problem 28

By leopinzon18 on Sat, 11/05/2016 - 10:02

what formulas are being used

By Mr. Dychko on Tue, 11/08/2016 - 00:42

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 - 19:43

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!

By Mr. Dychko on Tue, 11/08/2016 - 00:37

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 - 11:51

Why is the negative sign from the v2 prime early not considered in the final calculation. Thanks.

By Mr. Dychko on Tue, 11/08/2016 - 00:29

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 - 08:23

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?

By Mr. Dychko on Sat, 10/29/2016 - 13:01

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 - 11:35

Can you give a brief explanation of what you mean by "energy goes into heat"? Is it transformed into heat? Thanks.

By Mr. Dychko on Sat, 10/29/2016 - 11:49

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 - 18:46

Does anyone know why the 8 degree incline is ignored?

By Mr. Dychko on Sat, 10/29/2016 - 11:44

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 - 15:26

no sound ;-(

By Mr. Dychko on Sat, 10/29/2016 - 11:40

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 - 11:43

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

By Mr. Dychko on Sun, 10/16/2016 - 03:53

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 - 21:31

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.

By Mr. Dychko on Sun, 10/16/2016 - 03:59

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.

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 2, Problem 49

By idan on Thu, 10/13/2016 - 17:13

Do you think it's acceptable to use a graphing calculator to find the zeros of these functions? We're studying physics, not math.

By Mr. Dychko on Sun, 10/16/2016 - 03:47

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

Giancoli 7th Edition, Chapter 2, Problem 39

By lcbishop on Sat, 10/08/2016 - 17:17

If for instance we considered down not to be positive and ended up with a negative. Does this essentially make the answer wrong?

By Mr. Dychko on Sun, 10/09/2016 - 13:00

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".

Hope that helps,
Mr. Dychko