I solved the problem using only kinematics equations. Fnet = ma = mgsin30° - Nμ a = g*sin30° - g*cos30°*μ V^2 = 2*a*d = 2*(g*sin30° - g*cos30°*μ)*(h/sin30°) v = 12,74 m/sec
Why not account for friction by saying that friction does work on the system (which would be negative) rather than saying it's a change in energy due to heat lost?
Why is it that sometimes the work done by the friction force is at the initial side of the equation and other such as this is at the final side of the equation???????
Without knowing which "other" example you are referring to, the general answer is that it probably depends on what you are trying to determine. Here in this example we are using the equation that is used for most energy conservation problems. The energy put into the system + energy already in the system must equal the energy remaining in the system + energy removed from the system by the friction force.
This equation: W+ PEo+KEo= PEf+KEf + Heat lost, is this the formula for work done on a system by an external force, with friction? It's just that PEo+KEo=PEf+KEf is reminding me of something?
I agree with all his method, however notice that all his questions give the coefficient of friction, and unfortunately it is never stated if we're given static or kinetic coefficient of friction.
Regarding a scenario where the kinetic coefficient of friction is unknown, I'm assuming we'd consider forces and constant acceleration eqns to find the coefficient?
Is the force of friction supposed to be negative in the heat lost equation since it opposes the motion of the block's acceleration down the incline? Or does direction of forces not matter in energy questions?
Dear Gannu, Despite force being a vector quantity, work is scalar quantity. Therefore the measurement is scalar. The same difference between speed and velocity; speed refers to the magnitude of speed measure, velocity takes into account the direction of the movement. I hope this helps.
I got the same question. Could you please explain more details in this issue. Why we are no need to consider the work done by the parallel component of weight? Thanks!
I love your videos...the best in the web....... But, in this example you added work made for the friction force positive...It was not supposed to be negative since the force of friction is going opposite to the motion????
I have a similar problem where the height of the ramp as well as the distance traveled is unknown. I have the Friction Coef, the mass and the initial velocity. I assume the final velocity should be 0, because the question is 'what is the length of the ramp?' How do I solve for d without h?
Work is due to an outside force (not including gravity) that works on the system during the process. (Gravity is taken into account by the potential energy)
thank you so much for the videos, im very grateful, i have a doubt sir, since the friction force is at a 180 degree angle to the displacement shouldnt heat loss be Force(friction)*d*cos 180 = -fd
No, it is positive. The left side of the equation represents the energy you started with. The right side of the equation includes the energy you have left. If the energy you have left is less than the amount you started with (because you lost energy due to friction), then you must ADD the amount lost on the right side for the two sides of the equation to be equal.
It is worked the exact same way, and uses the exact same equation. If the box is pushed (is it continuous or a momentary impulse?) then you add the energy imparted on the object because of the work done by the force. (W = F * d)
AGAIN NEED A RESPONSE ASAP! Work done by Friction has to be negative because displacement and force of friction are in the opposite direction. F of k * d * cos (180)?? Am i right? I need a response. I got AN EXAM on Tuesday!!!
That is because no work was done on the system while the block was sliding down. (The work done by the friction force is accounted for by the energy lost term)
Does anyone in here get when i say that just memorizing a freaking equation won't actually make you understand the problem as a whole?! That's why i love chess, because i can actually learn the deep reason why i'm making such move, otherwise than physics taught in schools (at least in mine).
My left ear loved this.
god tier comment
Studying for a physics exam right now and your videos are perfect! I learn so much by working along with the video! Thanks for doing what you do!
the best lecture for physics on UA-cam I've ever seen.
You are really good at explaining in all your videos , with that you're helping me pass my physics class !
you are the best physics lecture i have ever seen
Leyla,
Thanks for the feedback. Much appreciated.
Alex Eul I don't like to make such "absolute" statements. However most of these types of problems can be solved with that equation.
I solved the problem using only kinematics equations.
Fnet = ma = mgsin30° - Nμ
a = g*sin30° - g*cos30°*μ
V^2 = 2*a*d = 2*(g*sin30° - g*cos30°*μ)*(h/sin30°)
v = 12,74 m/sec
Thanks for your lectures. I like the excitement you show on the geometry related parts.
Happy we can help.
best lecture ever, highly recommend
isn't there work done by the Weight vector component that goes along the slope?
Thank you so much. This problem was driving me crazy
Glad it helped
Your vids are the best!
thank you so much sir you're a great help! I like how you teach it was so handy to learn. learner here from Philippines
Welcome to the channel!
Very help full, thanks!
Glad to hear it!
Good work...your tutorials are of good help indeed......I didn't know we exclude work done by gravity on sliding motion...Thank you
how did you get the formula at the beginning of the video??
This is the energy conservation equation. The energy you have initially + any work put into the system must equal the final energy + any energy lost
@@MichelvanBiezen ah that makes a lot of sense, may I also ask where you got the 1 in the 1-costheta/sintheta?
Because of you Sir I am starting to love physics 👌 Thank you Sir☺
That's great. Thanks for sharing.
Michel van Biezen 👍👍
Why not account for friction by saying that friction does work on the system (which would be negative) rather than saying it's a change in energy due to heat lost?
That is a perfectly fine method as well.
@@MichelvanBiezen wasn't sure if it was conceptually incorrect while yours was correct. Thanks!
It is fine, but then you make it a negative value, you must place that term on the left side of the energy equation (instead of the right side).
Why is it that sometimes the work done by the friction force is at the initial side of the equation and other such as this is at the final side of the equation???????
Without knowing which "other" example you are referring to, the general answer is that it probably depends on what you are trying to determine. Here in this example we are using the equation that is used for most energy conservation problems. The energy put into the system + energy already in the system must equal the energy remaining in the system + energy removed from the system by the friction force.
why did you canceled the work in the equation when there is mgsin0 acting on the crate
The work done by gravity is already accounted for in the initial and final potential and kinetic energy (using this equation).
Why doesn't gravity acting on the crate count as work?
It does, but that work is accounted for by the "mgh" term
@@MichelvanBiezen ok thx
actually that cosine/sine can simplify to cot or 1/tan
Iam a srilankan Actually i don't know what are you saying but I can understand that idea.... You explain it very well sir thanks for it
Welcome to the channel!
Marvelous. Hiwever, I am surprised why mgsin(theta) wasnot used at all especially as an initial work on the system?
mg sin(theta) is accounted for in the PE term mgh. You cannot count it twice.
This equation: W+ PEo+KEo= PEf+KEf + Heat lost, is this the formula for work done on a system by an external force, with friction? It's just that PEo+KEo=PEf+KEf is reminding me of something?
That equation works for any situation. W = work done on the system adding energy and E lost is energy lost by the system, typically due to friction.
can we substitute width or length of ramp for H?
d represents the length of the ramp, but since we were not given the length of the ramp we had to find the length d in terms of h d = h / sin(theta)
@@MichelvanBiezen oh, so for height i could Dsin30 to get H
I agree with all his method, however notice that all his questions give the coefficient of friction, and unfortunately it is never stated if we're given static or kinetic coefficient of friction.
Since the objects are moving, it must be kinetic friction.
Regarding a scenario where the kinetic coefficient of friction is unknown, I'm assuming we'd consider forces and constant acceleration eqns to find the coefficient?
That is correct.
Thank you. You're quite prompt with your replies.
Is the force of friction supposed to be negative in the heat lost equation since it opposes the motion of the block's acceleration down the incline? Or does direction of forces not matter in energy questions?
Dear Gannu,
Despite force being a vector quantity, work is scalar quantity. Therefore the measurement is scalar. The same difference between speed and velocity; speed refers to the magnitude of speed measure, velocity takes into account the direction of the movement. I hope this helps.
Why does the parallel component of weight not contribute to the equation?
Its contribution is already accounted for by the change in KE and PE.
I got the same question. Could you please explain more details in this issue. Why we are no need to consider the work done by the parallel component of weight? Thanks!
sir , won't there be initial work due to force (mg sin tetha)..
I love your videos...the best in the web....... But, in this example you added work made for the friction force positive...It was not supposed to be negative since the force of friction is going opposite to the motion????
I have a similar problem where the height of the ramp as well as the distance traveled is unknown. I have the Friction Coef, the mass and the initial velocity. I assume the final velocity should be 0, because the question is 'what is the length of the ramp?' How do I solve for d without h?
h = d * sin(theta). If the angle is known then you only have one unknown
I don't get the part with the factorization. Why does it say [1 - blablubb] ???? The 1 - confuses me. Can somebody explain that please.
+ProAiming1337
When you factor out 2gh from each of the last 2 terms what remains is ( 1 - (cos (theta) mu / sin(theta))
Is the equation applicable to all problems involving work, energy, and power?
Yes. But there are some problems for which you'll need additional equation(s).
thanl you so much sir great work
Thank you. Glad you find it helpful. 🙂
Why can't we equate the friction force with mgsin(theta) instead of using the friction force formula?
To determine the friction force we must use the strict definition: Friction force = normal force x coefficient of friction
why not work done by friction is negative in this case ??????
It depends on your reference. Strictly speaking W = F x d x cos(theta) and thus you will get a negative answer.
left ear enjoyed well....
all jokes aside, thank you sir for the help
Glad it was helfpul! 😂🙂😂
@@MichelvanBiezen Lol!
there is an easyer way . you can figure out d useing sin 30 which is 0.5 . d is 20m
thanks sir..!!
Why work is zero??
Work is due to an outside force (not including gravity) that works on the system during the process. (Gravity is taken into account by the potential energy)
Sir why we didn't consider work done by mgsin(theta) here
That is accounted for with the change in potential energy.
@@MichelvanBiezen sir! Why we cant consider its psuedo force and put the work done by it in that equation
thank you so much for the videos, im very grateful, i have a doubt sir, since the friction force is at a 180 degree angle to the displacement shouldnt heat loss be Force(friction)*d*cos 180 = -fd
No, it is positive. The left side of the equation represents the energy you started with. The right side of the equation includes the energy you have left. If the energy you have left is less than the amount you started with (because you lost energy due to friction), then you must ADD the amount lost on the right side for the two sides of the equation to be equal.
what if the box was pushed from the top ? what can i do then? and they want the final velocity to = 0 m/s ....
It is worked the exact same way, and uses the exact same equation. If the box is pushed (is it continuous or a momentary impulse?) then you add the energy imparted on the object because of the work done by the force. (W = F * d)
Why is the cos(thetre) =30 degrees? Shouldn't it be 180 because it's the normal force
Sheedo Bulbulia
The angle theta is the angle between the normal to the plane and the vertical
This is interesting
Glad you liked it.
AGAIN NEED A RESPONSE ASAP! Work done by Friction has to be negative because displacement and force of friction are in the opposite direction. F of k * d * cos (180)?? Am i right? I need a response. I got AN EXAM on Tuesday!!!
+Sai Thu See my response on your previous question.
what if i add a force F to the object how can i find it's F work
Brandon,
There are other videos with that example.
why didn't you use mgSin(theta)
+Mulanga Mphagi It was not needed here, since we didn't have to calculate the acceleration and it doesn't contribute to the friction force.
@@MichelvanBiezen ok sir! Then it is just contributing in potential energy but no need to write in equation maybe!
Great! :)
Thanks!
@@MichelvanBiezen You bet! :)
You just dropped the W?
That is because no work was done on the system while the block was sliding down. (The work done by the friction force is accounted for by the energy lost term)
Michel van Biezen Oh okay. Thanks for the clear up, and all your videos. They are extremely helpful!
How about mgsin seta
That is accounted for by the change in potential energy
Does anyone in here get when i say that just memorizing a freaking equation won't actually make you understand the problem as a whole?! That's why i love chess, because i can actually learn the deep reason why i'm making such move, otherwise than physics taught in schools (at least in mine).
the only part I didn't get was where 1/2 mv squared came from
+nikotwenty (1/2) m v^2 is the kinetic energy gained by the block sliding down the incline
+Michel van Biezen OK I get it, I looked up formula for kinetic energy and that was the base formula. I didn't know that before
+nikotwenty Isn't this guy THE bae???
nvm im wrong
i'm subscribing... definitely