I can't find a word to describe you. God bless you Professor. Keep going. You changed my viewpoint to Physics. I used to hate. But now I love it because you explained well. THANK YOU
+Michel van Biezen I believe that most people would find physics very fun and interesting if they got a fair chance to discover it. For some reason a lot of people seem to lack confidence when it comes to mathematical subjects in general, and I am not sure why - but I do know that those people can develop a great interest in those topics if they discover how useful they can be, and how they can apply it. And of course, they obviously need to gain a solid understanding for Arithmetic and Algebra, since those things are the foundation for literally everything that comes later.
Michel is great because he clearly explains the concepts. Some teachers seem to be content with throwing around formulas everywhere, and that's a big mistake - one rule of thumb in math and physics is to always try to understand the concepts so well that you don't need to look up any formulas.
A 15kg box is pushed up an incline 30° to the horizontal and 3.3m long with a speed of 6ms-¹ at the bottom. Due to friction, the box stopped 2.0m and slides back down. a) What is the magnitude of the friction acting on the box assuming it is constant. b) How fast is the box moving when it slid back to the bottom. I solved with the energy formula and I got magnitude of friction as 60 N. With F = uR, I got 75 where u = tan30 and R = mgcos 30
Thanks for your quick response Mr Biezen. This is the problem - my solution is included. I am not sure if i am sure it would be prudent to ask specific question here, but i will give it a try: A block of mass 90 kg sits on a slope with an angle of inclination 12° above the horizontal. The coefficients of friction between the block and the slope are µk = 0.03 and µs = 0.2. a. What is the maximum frictional force that will allow the block to stay in place? b. At what angle of inclination will the block start to slide down the slope? Solution for part (a) F_s(max.) = µsN F_s(max.) = µsmgcosθ F_s(max.) = 0.2(90kg)(9.81m/s^2 )(cos12) F_s(max.) =172.72N Maximum frictional force that will allow the block to stay in plaxe = 172.72N Solution for part (b) fg = fs mgsinθ = µsN mgsinθ = µsmgcosθ sinθ = µscosθ µs = sinθ/cosθ= tanθ θ = tan^(-1)(µs) θ = tan^(-1)(0.2) θ = 11.31° The block will start to slide down the slope at 11.31° I am confused because the block is expected to slide at 11.31 or an angle greater, but yet, 12 degrees in part (a) is suggesting the block is at rest? Is my solution for part (a) correct? I am thinking i should have found part b first, then use 11.31 degrees and enter it in the formula: F_s(max.) = µsmgcosθ , instead of using 12 degrees, use 11.31 degrees.
You are correct. In order for the block to be stationary at an angle of 12 degrees the coefficient of static friction should be 0.213. Therefore the block will slide when the angle is 12 degrees.
Assuming that KE_0 and KE_f are 0 is an unnecessary assumption. Since the box isn't accelerating, the initial and final kinetic energies are equal, so they just cancel each other in your equation.
jampk24 thanks man I was confused on that I was like how can KEfinal be 0 if KE is 0.5mv^2? Now I know it is because the box wasn’t accelerating and the KEinitial was equal to KEfinal. Thank yih
Thank you for your video. I would like to mention that your videos are unique and completely different from the others. But, the question is about the KE0. How we could consider the V0 = 0 while already knew that the mass was moving upwards with a constant speed. I mean, if V0(initial) = 0 then: A: How the mass will go up with out any speed? B: Then some acceleration would be needed for taking a constant speed more than 0. (Am I right?) I would like to put forward that we could have considered KE0 = 0.5m(V0)^2 and later on we could have considered KEf = 0.5m(Vf)^2 and as the question says, V0=Vf=C (Constant) and it means that KE0 = KEf = 0.5m(C)^2 and they will cancel out of our equation. Sorry, for leaving too much comments. I would say that I like physics problems and discussing about the solutions.
+Michel van Biezen Thank you for your reply. What I'm saying is that if we assume the velocity equal to 0 at start point as you mentioned, so our mass should take some more velocity to move. It's clear that with 0 velocity, our mass won't move at all at if there is any increase in the velocity later, it is in contrary with the question's given information. Could you make me understood about it? Thanks for your attention.
+Daniyal Ravandeh You see, there are two KE equations originally. Even if you substitute any value of V, they would still cancel out. Since the mass and velocity don't change, KE1 - KE2 = 0, because 1/2*m1*v1^2 - 1/2*m2*v2^2 = 0, where m1 = m2 and v1 = v2.
In order for the box to move at constant v, it has to over come 2 forces; mgsin(th) and friction and all of that multiplied by 100 which gavee me 6605, so why is our answers different. I know you are right, but I dont know where I went wrong?
Quick question. Couldn't you also solve for work by finding the force as (mgsintheta)costheta and subtract the force of friction, then multiply everything by the distance? I saw in your last video(without friction), you found the force as mgsintheta then multiplied by D. How come you can't just do the same thing here, except use FDcostheta and subtract friction from this? Sorry if I was a bit confusing.
I am a little bit confused... shouldn't the net work by calculated by finding the work done by gravity and subtract it from the work done by friction? I found your answer when I added them though. But What if I subtract them? my thinking is energy is lost due to having a nonconservative force( friction) involved. The total work performed should the difference right?
What you are suggesting is the net work if the box were sliding down the ramp; in that case, work would be done by gravity. In this problem, you are effectively doing work against gravity. The net work is the total work needed to lift the box up by a height h AND also overcome friction. Addition should feel like the logical choice.
I certainly enjoy your videos. I have a question based on this video. If a block is at rest on the incline plane at an angle of 30 degrees, would the maximum frictional force that would allow the block to stay in place be calculated using 30 degrees, OR would we have to find the minimum angle of the incline plane needed for the block to slide before we can determine the maximum frictional force? That is, find that minimum angle and then substitute it in the formula for fs(max) = UsMgcos (minimum angle needed for block to slide)?
It depends on how the question is worded. Sometimes they will ask what coefficient of friction is required to keep a block from sliding for a given angle. Sometimes they ask: for a given coefficient of friction, what is the maximum angle you can have before the block begins to slide?
If both kinetic energies are 0, meaning both initial and final velocities (in the 1/2 mv^2 formula for K.Eo and K.E1) of the box are 0. So the box literally moved up the plane without velocity. Practically speaking, the box moved up the inclined plane without moving.
Potential Energy Final equals mgyf, why is it mgd for the final equation. It should be multiplied by the height which is 25 meters not by the overall displacement of 50 meters. My final answer was 4937 Joules
i find it inconsistent that in physics Kg is standard, i always want to write 1000 instead of 1. Why all other units are prefix-less except this one is annoying.
The result in the video is correct. Don't automatically assume "something" is an addition or a subtraction, it depends on how the equation is formulated.
Forces are not automatically positive or negative. A term in an equation will be positive or negative depending on the meaning of the term and its position in the equation. For example: x + 5 = 3 (x = -2) 5 = 3 + x (x = 2). The energy equation equates the initial energy to the final energy of the block. In order for that equation to be correct you must ADD energy lost due to friction on the right side.
Energy is not conserved when there is friction. But if we account for all energy gained and all energy lost, we can set up an equation that is balanced.
Since the velocity didn't change, the kinetic energy didn't change so we can just set the KE before and after equal to zero for simplicty. You can plug in any value, but since we were not told what the speed was, it was just easier to call it zero.
Good sir, i'm wondering the difference between this and video136 in this playlist is considering in the other video kinetic energy isn't 0 although it seems exactly the same. Also the applications of energy was given by a different equation. Thank You
NEED A RESPONSE ASAP!! Word done by friction force should be (-) negative? Since friction and displacement are in the opposite direction which would be cos(180), F of k * d * cos (180)
+Sai Thu F net = m total * aAssume a direction for the acceleration.Then F net = All the forces aiding the acceleration (positive) - all the forces opposing the acceleration (negative) Friction is an opposing force therefore it is negative
+Michel van Biezen Thx for quick response. I understand the part u said. @ 7:10, for work done by friction force, F of k * d, should it not be F of k * d * cos (180)? Because displacement is going to other direction which is the opposite of force of friction according to Work theory W = F d cos ( @ ).
+Sai Thu You are letting your intuition get in the way of understanding. Look at each step in the video carefully and make sure you understand each step. You will see that each step makes sense. In the end the work put into the system must be equal to the potential energy gained + the work done to overcome friction.
If you place the heat lost on the right side of the energy conservation equation (as shown in the video), the energy lost must be positive, otherwise the equation is not balanced.
The definition of the friction force = N * u (which is the normal force multiplied by the coefficient of friction). The normal force on an inclined plane = mg cos(theta)
Biezen, shouldn't we take in consideration to the gravite x-composant`? A box with 50 kg mass is pushed up by a incline plane. Hyp = 2m , h=1m. Fr= 50N. I did like this at the lesson: Work needed = mgh + Fr*d But my teacher said something about the force of the x-composant. How would i solve the problem by using that, would i get the same result? :) Since i watched this video i learned the way i did, but now im unsure.!
+Jonasa Karlsson There are two ways to solve for work done: 1) Multiply the force component in the direction of travel with the distance traveled. 2) Multiply the force with the distance traveled times the cosine of the angle between the two. (Actually these two methods are the same but in different format)
+Michel van Biezen so if Fr was given=50N Work needed to push a box a distance d of a hight of h , with mass m : W_needed= mgh + 50N*d Thats, the way you solves the problem here. But if i would take in consideration the x-composant of the gravity not the y-composant, since the friction "Fr" is given as 50N there is no coefficent. How would one solve it? You need to push with a force that overcome the friction force , 50N + overcome mg *sin (theta), and all that in a parantese multiplied with the distance or how would the second way be to solve it, wouldn't you need to take in consdieration the hight? Now when i think about it , the hight is related to distance and sinus, so maybe "mgh" is just a part of the "x-composant of gravity"?
why doesn't it have kinetic energy even though it states in the question it moves slowly. is it because it moves in constant speed. meaning there is no velocity hence we can't calculate the kinetic energy 1/2mv^2?
Another thing, since the speed is constant, v before equals to v after. Since mass is constant, one Ke is positive and another is negative, they cancel out no matter how fast the speed was
I hope to have a professor like you. You explained it really really well while my prof just gave us the lesson online without any example at all then gave us the activities with a confusing questions. What l mean by that is the questions are incomplete and the composition is a bit confusing also the work done are so impossible. I JUST WANNA CRY RIGHT NOW😭😭. I never had a hard time answering physicsn but right now? I juat don't understand it at all. I desperately want to understamd the concept but her questions are so hard to understand that l can't even find it in the internet.
i am bit surprised. It was superb technique to solve the questions that way, however work is define as f.dcos(teta) work done by the kinetic friction is f.dcos(teta 180 degree which implies us energy transfer from block to force parallel to the surface oppose the motion of block but you neglected 180 degree made cos(0) which gives us positive result. Why you didn't suppose to be subtracted work done by the kinetic frictional force.
Why is the kinetic energy 0? Although the object is not accelerating, it still has a velocity. Thus, if it has velocity, shouldn't it have kinetic energy?
We only need to keep track of a change in kinetic energy and since the speed is constant, no work was done to increase the kinetic energy (the increase was 0)
why didnt you include the mg sin (theta) in the friction as well?? i am really confused when to include it in the calculations. could you please help me with it?
The definition of the friction force is: Ffriction = Normal force x coefficient of friction. Sine mg sin(theta) does not contribute to the normal force (since it is parallel to the slanted surface) it does not contribute to the friction force.
mass 1250 kg moves from the bottom to the top of a hill of length 500m amd height 30m. power of engine is constant and equal to 30kW. a= 4m/s^2 at bottom a= 2 m/s^2 at top resistance to motion = 1000N. a) find the car's gain in kinetic energy. b) the work done by the cars engine.. could you please solve it.
Is there another way to solve that problem? I tried with the method you gave in the previous video, but it didn't workout, even though (to me atleast) it's almost the same problem.
The method used to solve the problem, depends on the conditions. In this case, the box moves upward at constant speed. I suggest you use the method shown in the video.
It is true that with projectile motion, sin is usually related to the vertical motion and cosine is usually related to the horizontal motion, but assuming that it is always that way, can lead to incorrect answers. That is why it is a good idea to draw out the triangle in your sketch and confirm the correct trig function. In the case of an inclined plain the component of the weight (mg) of the object perpendicular (normal) to the incline can be found using the cosine of the angle.
The frictional force should be negative because the angle between μN and displacement is 180 degrees and both the vectors are in opposite direction. I'm really confused, please help me professor.
I understand why you are confused.When we work applications of Newton's second law types of problems (F = ma) then yes, the friction force is "negative" or "in the opposite direction of the motion without friction. But in this type of problem, we are equating the left side (which has all the initial energy + energy added to the system) to the right side (which has all the final energy + energy removed from the system by friction). That is why we add the energy lost on the right side of the equation.
I got it professor! Thank you so much for making these videos. I have shared this channel to all of my friends and literally they are so happy and thankful to you as well as me coz I shared this wonderful channel with them. Your videos helped me a lot. Tons of blessings from my side.
I am dealing with the same type of question, but my problem is the question in the video have four given elements, while the question I am dealing with has only three given elements (displacement is unknown), is there an alternative way to deal with this type of question?
There are different ways to solve a problem. Try it that way and see if you get the same answer, (and by doing so, you'll get the answer to your question).
You can try and see of you get the same answer. Although for this particular problem, the conservation of energy technique shown in this video is probably easier.
can someone help me out. I understand that if the box starts at rest then the initial vel would be zero which makes initial KE zero but how is the final KE also zero if it has a speed? even if the speed does not change there has to be some speed to get the box up the incline?
we are not worrying about significant figures here, just the concepts and how to find the answers. Significant figures are covered in a different playlist. We are not assuming that the numbers given are "measured" values. In the case where we are actually using measured values we should then use the principles of significant figures.
If for example, the speed was increasing, then part of the work would be converted to kinetic energy and part of the work would be converted to potential energy minos any energy lost due to friction.
It depends on which side of the equation that term is placed. If you place the "energy lost" term on the left side it becomes negative and if you place it on the right side it becomes positive.
Why using the equation "(F - friction) * distance" has different outcome? Isn't it about 3206? This is from that you plus "cos30*0.2" rather than minus them although the "cos30*0.2" is from the friction that is the opposite direction from the force. Could you explain it to me?
The force is pushing the block up the hill, in the same direction as the motion. The friction force pushes in the opposite direction of the motion of the block, therefore it does negative work on the block.
@@MichelvanBiezen Oh, the reason why you plus the heat loss rather than minus the heat loss is because we want to figure out how much amplitude we need to move the box. Am I right?
I can't find a word to describe you. God bless you Professor. Keep going. You changed my viewpoint to Physics. I used to hate. But now I love it because you explained well. THANK YOU
+Siamantoo Zohrabiyan Thanks for writing. I am glad you like physics now.
+Michel van Biezen
I believe that most people would find physics very fun and interesting if they got a fair chance to discover it.
For some reason a lot of people seem to lack confidence when it comes to mathematical subjects in general, and I am not sure why - but I do know that those people can develop a great interest in those topics if they discover how useful they can be, and how they can apply it.
And of course, they obviously need to gain a solid understanding for Arithmetic and Algebra, since those things are the foundation for literally everything that comes later.
Thanks ...you know I actually prefer seeing you write on the board than some digitalised video where I just hear you voice.. you made physics rock !!♡
i used to hate physics so much... I CANT BELIEVE YOU CHANGED THIS IN TWO DAYS, I FEEL MUCH MORE CONFIDENT IN PHYSICS NOW. thanks Prof. Michel!!
agreed. He is great!
Michel is great because he clearly explains the concepts.
Some teachers seem to be content with throwing around formulas everywhere, and that's a big mistake - one rule of thumb in math and physics is to always try to understand the concepts so well that you don't need to look up any formulas.
Glad to hear it :)
A 15kg box is pushed up an incline 30° to the horizontal and 3.3m long with a speed of 6ms-¹ at the bottom. Due to friction, the box stopped 2.0m and slides back down.
a) What is the magnitude of the friction acting on the box assuming it is constant.
b) How fast is the box moving when it slid back to the bottom.
I solved with the energy formula and I got magnitude of friction as 60 N.
With F = uR, I got 75 where u = tan30 and R = mgcos 30
If you assume g = 10 m/sec^2 , then the friction force is 60 N
Thanks for your quick response Mr Biezen. This is the problem - my solution is included.
I am not sure if i am sure it would be prudent to ask specific question here, but i will give it a try:
A block of mass 90 kg sits on a slope with an angle of inclination 12° above the horizontal. The coefficients of friction between the block and the slope are µk = 0.03 and µs = 0.2.
a. What is the maximum frictional force that will allow the block to stay in place?
b. At what angle of inclination will the block start to slide down the slope?
Solution for part (a)
F_s(max.) = µsN
F_s(max.) = µsmgcosθ
F_s(max.) = 0.2(90kg)(9.81m/s^2 )(cos12)
F_s(max.) =172.72N
Maximum frictional force that will allow the block to stay in plaxe = 172.72N
Solution for part (b)
fg = fs
mgsinθ = µsN
mgsinθ = µsmgcosθ
sinθ = µscosθ
µs = sinθ/cosθ= tanθ
θ = tan^(-1)(µs)
θ = tan^(-1)(0.2)
θ = 11.31°
The block will start to slide down the slope at 11.31°
I am confused because the block is expected to slide at 11.31 or an angle greater, but yet, 12 degrees in part (a) is suggesting the block is at rest? Is my solution for part (a) correct? I am thinking i should have found part b first, then use 11.31 degrees and enter it in the formula: F_s(max.) = µsmgcosθ
, instead of using 12 degrees, use 11.31 degrees.
You are correct. In order for the block to be stationary at an angle of 12 degrees the coefficient of static friction should be 0.213. Therefore the block will slide when the angle is 12 degrees.
Assuming that KE_0 and KE_f are 0 is an unnecessary assumption. Since the box isn't accelerating, the initial and final kinetic energies are equal, so they just cancel each other in your equation.
jampk24 thanks man I was confused on that I was like how can KEfinal be 0 if KE is 0.5mv^2? Now I know it is because the box wasn’t accelerating and the KEinitial was equal to KEfinal. Thank yih
Sir, you are the greatest. I would love to work under your supervision.
I just wanna say thank you for helping me get A in my physics class.
Great job! (Glad we could help)
I think I need to work more on this.
Thank you for your video. I would like to mention that your videos are unique and completely different from the others.
But, the question is about the KE0. How we could consider the V0 = 0 while already knew that the mass was moving upwards with a constant speed.
I mean, if V0(initial) = 0 then:
A: How the mass will go up with out any speed?
B: Then some acceleration would be needed for taking a constant speed more than 0. (Am I right?)
I would like to put forward that we could have considered KE0 = 0.5m(V0)^2 and later on we could have considered KEf = 0.5m(Vf)^2 and as the question says, V0=Vf=C (Constant) and it means that KE0 = KEf = 0.5m(C)^2 and they will cancel out of our equation.
Sorry, for leaving too much comments. I would say that I like physics problems and discussing about the solutions.
+Daniyal Ravandeh
Can you boil this down to a single question?
+Michel van Biezen
Sorry for asking my question in a confusing way.
Could you explain why you have considered KE0 = 0 ???
+Daniyal Ravandeh KE energy depends on the velocity: KE = (1/2) m v^2In the beginning we can assume that v = 0 therefore there will not be any KE
+Michel van Biezen Thank you for your reply.
What I'm saying is that if we assume the velocity equal to 0 at start point as you mentioned, so our mass should take some more velocity to move. It's clear that with 0 velocity, our mass won't move at all at if there is any increase in the velocity later, it is in contrary with the question's given information. Could you make me understood about it?
Thanks for your attention.
+Daniyal Ravandeh
You see, there are two KE equations originally.
Even if you substitute any value of V, they would still cancel out. Since the mass and velocity don't change, KE1 - KE2 = 0, because 1/2*m1*v1^2 - 1/2*m2*v2^2 = 0, where m1 = m2 and v1 = v2.
In order for the box to move at constant v, it has to over come 2 forces; mgsin(th) and friction and all of that multiplied by 100 which gavee me 6605, so why is our answers different. I know you are right, but I dont know where I went wrong?
Very good explanation dear
Thank you so much 🙂
Quick question. Couldn't you also solve for work by finding the force as (mgsintheta)costheta and subtract the force of friction, then multiply everything by the distance? I saw in your last video(without friction), you found the force as mgsintheta then multiplied by D. How come you can't just do the same thing here, except use FDcostheta and subtract friction from this? Sorry if I was a bit confusing.
There are often multiple ways to solve a problem, and the best way to learn is to try it in a different way and see if you get the same answer.
I am a little bit confused... shouldn't the net work by calculated by finding the work done by gravity and subtract it from
the work done by friction? I found your answer when I added them though. But What if I subtract them? my thinking is
energy is lost due to having a nonconservative force( friction) involved. The total work performed should the difference
right?
The method and answer in the video is correct.
What you are suggesting is the net work if the box were sliding down the ramp; in that case, work would be done by gravity. In this problem, you are effectively doing work against gravity. The net work is the total work needed to lift the box up by a height h AND also overcome friction. Addition should feel like the logical choice.
I certainly enjoy your videos. I have a question based on this video. If a block is at rest on the incline plane at an angle of 30 degrees, would the maximum frictional force that would allow the block to stay in place be calculated using 30 degrees, OR would we have to find the minimum angle of the incline plane needed for the block to slide before we can determine the maximum frictional force? That is, find that minimum angle and then substitute it in the formula for fs(max) = UsMgcos (minimum angle needed for block to slide)?
It depends on how the question is worded. Sometimes they will ask what coefficient of friction is required to keep a block from sliding for a given angle. Sometimes they ask: for a given coefficient of friction, what is the maximum angle you can have before the block begins to slide?
If both kinetic energies are 0, meaning both initial and final velocities (in the 1/2 mv^2 formula for K.Eo and K.E1) of the box are 0.
So the box literally moved up the plane without velocity.
Practically speaking, the box moved up the inclined plane without moving.
If the velocity did not change, then the kinetic energy on both sides of the equation will cancel out. (v is not zero).
@@MichelvanBiezen Ok, I see. Thanks for the feedback
Potential Energy Final equals mgyf, why is it mgd for the final equation. It should be multiplied by the height which is 25 meters not by the overall displacement of 50 meters. My final answer was 4937 Joules
h is replaced by d sin (theta). Also, work must be done to overcome friction, thus the final answer is greater.
i find it inconsistent that in physics Kg is standard, i always want to write 1000 instead of 1. Why all other units are prefix-less except this one is annoying.
Sir I love your way of teaching.Thank you sir
would it be easier pulling the box? instead of pushing it?
Hello Mr there is a fraction the answer supposed to be 2802.6J because it’s substirction not addition .
The result in the video is correct. Don't automatically assume "something" is an addition or a subtraction, it depends on how the equation is formulated.
Why didn't you use cos180 when writing the formula for friction? Isn't friction supposed to be negative?
Forces are not automatically positive or negative. A term in an equation will be positive or negative depending on the meaning of the term and its position in the equation. For example: x + 5 = 3 (x = -2) 5 = 3 + x (x = 2). The energy equation equates the initial energy to the final energy of the block. In order for that equation to be correct you must ADD energy lost due to friction on the right side.
why there is consrevation of energy ? we have friction and its non conservative force so why there is conservation of energy ?
Energy is not conserved when there is friction. But if we account for all energy gained and all energy lost, we can set up an equation that is balanced.
A question, how is final kinetic energy 0? I'm searching and all the answers say constant speed/velocity has a constant kinetic energy.
Since the velocity didn't change, the kinetic energy didn't change so we can just set the KE before and after equal to zero for simplicty. You can plug in any value, but since we were not told what the speed was, it was just easier to call it zero.
Good sir, i'm wondering the difference between this and video136 in this playlist is considering in the other video kinetic energy isn't 0 although it seems exactly the same. Also the applications of energy was given by a different equation. Thank You
My inclination is that because the other video has given force of 100N, then it is not a constant speed?
It depends. If the 100 N force is greater than mg sin(theta) + friction force, then the block will accelerate up the incline.
@@MichelvanBiezen Oops I was neglecting the weight components! Thank You for the quick response!
NEED A RESPONSE ASAP!! Word done by friction force should be (-) negative? Since friction and displacement are in the opposite direction which would be cos(180), F of k * d * cos (180)
+Sai Thu F net = m total * aAssume a direction for the acceleration.Then F net = All the forces aiding the acceleration (positive) - all the forces opposing the acceleration (negative) Friction is an opposing force therefore it is negative
+Michel van Biezen Thx for quick response. I understand the part u said. @ 7:10, for work done by friction force, F of k * d, should it not be F of k * d * cos (180)? Because displacement is going to other direction which is the opposite of force of friction according to Work theory W = F d cos ( @ ).
PLEASE HELP! This is bugging me.
Michel van Biezen Please HELP, Mr. Biezen!!
+Sai Thu
You are letting your intuition get in the way of understanding. Look at each step in the video carefully and make sure you understand each step. You will see that each step makes sense. In the end the work put into the system must be equal to the potential energy gained + the work done to overcome friction.
Shouldn't work done by friction be negative?
If you place the heat lost on the right side of the energy conservation equation (as shown in the video), the energy lost must be positive, otherwise the equation is not balanced.
u r so great professor .
why didn't you use mgSin(theta) since it's pushing up opposing the mgsin(theta) going downwards
The definition of the friction force = N * u (which is the normal force multiplied by the coefficient of friction). The normal force on an inclined plane = mg cos(theta)
Biezen, shouldn't we take in consideration to the gravite x-composant`?
A box with 50 kg mass is pushed up by a incline plane. Hyp = 2m , h=1m. Fr= 50N. I did like this at the lesson:
Work needed = mgh + Fr*d
But my teacher said something about the force of the x-composant. How would i solve the problem by using that, would i get the same result? :)
Since i watched this video i learned the way i did, but now im unsure.!
+Jonasa Karlsson
There are two ways to solve for work done:
1) Multiply the force component in the direction of travel with the distance traveled.
2) Multiply the force with the distance traveled times the cosine of the angle between the two.
(Actually these two methods are the same but in different format)
+Michel van Biezen so if Fr was given=50N
Work needed to push a box a distance d of a hight of h , with mass m :
W_needed= mgh + 50N*d
Thats, the way you solves the problem here.
But if i would take in consideration the x-composant of the gravity not the y-composant, since the friction "Fr" is given as 50N there is no coefficent. How would one solve it?
You need to push with a force that overcome the friction force , 50N + overcome mg *sin (theta), and all that in a parantese multiplied with the distance or how would the second way be to solve it, wouldn't you need to take in consdieration the hight? Now when i think about it , the hight is related to distance and sinus, so maybe "mgh" is just a part of the "x-composant of gravity"?
Jonasa Karlsson huh
Why kinetic energy final is 0 even when it has a slow but constant speed I believe change in k.e should be 0.
You are correct. However for this example, the word "slowly" is used such that the KE is negligible and can be ignored.
Michel van Biezen ok
Michel van Biezen Because V=constant, KEo =KEf, then the term KE cancel out...is it correct?
Thank you sir 👍👍👍
Most welcome
Thank you! So helpful.
why doesn't it have kinetic energy even though it states in the question it moves slowly. is it because it moves in constant speed. meaning there is no velocity hence we can't calculate the kinetic energy 1/2mv^2?
If v is small, we can ignore the KE (which is the purpose of this example)
oh.. i see. alright..
Another thing, since the speed is constant, v before equals to v after. Since mass is constant, one Ke is positive and another is negative, they cancel out no matter how fast the speed was
I hope to have a professor like you. You explained it really really well while my prof just gave us the lesson online without any example at all then gave us the activities with a confusing questions. What l mean by that is the questions are incomplete and the composition is a bit confusing also the work done are so impossible. I JUST WANNA CRY RIGHT NOW😭😭. I never had a hard time answering physicsn but right now? I juat don't understand it at all. I desperately want to understamd the concept but her questions are so hard to understand that l can't even find it in the internet.
We have over 2000 videos on physics on this channel. Are none of them related to your assignment?
Yup l just found it. Thanks a lot.
Btw what if the force is provided?
The method used is typically the same and carefully explained through all the examples.
i am bit surprised. It was superb technique to solve the questions that way, however work is define as f.dcos(teta) work done by the kinetic friction is f.dcos(teta 180 degree which implies us energy transfer from block to force parallel to the surface oppose the motion of block but you neglected 180 degree made cos(0) which gives us positive result. Why you didn't suppose to be subtracted work done by the kinetic frictional force.
We did. (It is the "heat lost" term). For the equation to be balanced Eo = Ef the energy lost must be added to the right side.
Why is the kinetic energy 0? Although the object is not accelerating, it still has a velocity. Thus, if it has velocity, shouldn't it have kinetic energy?
We only need to keep track of a change in kinetic energy and since the speed is constant, no work was done to increase the kinetic energy (the increase was 0)
@@MichelvanBiezen Ah!! Okay, thank you so much!!
why didnt you include the mg sin (theta) in the friction as well??
i am really confused when to include it in the calculations.
could you please help me with it?
The definition of the friction force is: Ffriction = Normal force x coefficient of friction. Sine mg sin(theta) does not contribute to the normal force (since it is parallel to the slanted surface) it does not contribute to the friction force.
mass 1250 kg
moves from the bottom to the top of a hill of length 500m amd height 30m.
power of engine is constant and equal to 30kW.
a= 4m/s^2 at bottom
a= 2 m/s^2 at top
resistance to motion = 1000N.
a) find the car's gain in kinetic energy.
b) the work done by the cars engine..
could you please solve it.
Is there another way to solve that problem? I tried with the method you gave in the previous video, but it didn't workout, even though (to me atleast) it's almost the same problem.
The method used to solve the problem, depends on the conditions. In this case, the box moves upward at constant speed. I suggest you use the method shown in the video.
If I want to calculate W=F*d, d=50m and F= mgsinθ+mgcosθμ ,is it correct?
why did you use mgcos(theta) for the normal force? shouldn't it be mgsin(theta) since sin is the y-component used for rise or fall?
It is true that with projectile motion, sin is usually related to the vertical motion and cosine is usually related to the horizontal motion, but assuming that it is always that way, can lead to incorrect answers. That is why it is a good idea to draw out the triangle in your sketch and confirm the correct trig function. In the case of an inclined plain the component of the weight (mg) of the object perpendicular (normal) to the incline can be found using the cosine of the angle.
I hate physics.... But because of you i changed my opinion completely...
Thank you, that was a great tutorial, I am wondering though why h isn't cos(30)*d, why is it sin(30)*d ?
You'll get the range or the distance (X) of the triangle instead of the height
How come that your acceleration is always 9.8m/s^2?
that's the gravitational acceleration
(20)*(9.81)*(50)*[sin(30)+cos(30)*(0.2)] = 6604 J ????
The frictional force should be negative because the angle between μN and displacement is 180 degrees and both the vectors are in opposite direction. I'm really confused, please help me professor.
I understand why you are confused.When we work applications of Newton's second law types of problems (F = ma) then yes, the friction force is "negative" or "in the opposite direction of the motion without friction. But in this type of problem, we are equating the left side (which has all the initial energy + energy added to the system) to the right side (which has all the final energy + energy removed from the system by friction). That is why we add the energy lost on the right side of the equation.
I got it professor! Thank you so much for making these videos. I have shared this channel to all of my friends and literally they are so happy and thankful to you as well as me coz I shared this wonderful channel with them. Your videos helped me a lot. Tons of blessings from my side.
I am dealing with the same type of question, but my problem is the question in the video have four given elements, while the question I am dealing with has only three given elements (displacement is unknown), is there an alternative way to deal with this type of question?
why didnt you use w=F*d?and F=mgsin theta?
There are different ways to solve a problem. Try it that way and see if you get the same answer, (and by doing so, you'll get the answer to your question).
sir can i use the other way you taught us from the previous video?
You can try and see of you get the same answer. Although for this particular problem, the conservation of energy technique shown in this video is probably easier.
the thumbnail is misleading. You have a pushing force acting in the x direction in the thumbnail and in this the force is going up the incline
You are correct. The force in the thumbnail should act parallel to the inclined plane.
Fixed it! Thanks!
can someone help me out. I understand that if the box starts at rest then the initial vel would be zero which makes initial KE zero but how is the final KE also zero if it has a speed? even if the speed does not change there has to be some speed to get the box up the incline?
+Chen Garces
If the kinetic energy doesn't change, it is the same in the beginning as it is in the end, then you can ignore it.
Why 2 sf, 20 has 1 sf
we are not worrying about significant figures here, just the concepts and how to find the answers. Significant figures are covered in a different playlist. We are not assuming that the numbers given are "measured" values. In the case where we are actually using measured values we should then use the principles of significant figures.
@@MichelvanBiezen by the way, what is the speed wasn't constant, what would we do, and what would be given?
If for example, the speed was increasing, then part of the work would be converted to kinetic energy and part of the work would be converted to potential energy minos any energy lost due to friction.
Can simplify answer W~ 6.6 kJ
What the Friction!! ...does
🙃
Devin Weston??
GTA????
everything is great except for the fact that the voice only comes from the right ear
Yes our older videos were filmed in mono sound (not stereo).
Isn't that friction oppose the motion ... Coz if that's case , work done by friction should be negative 1697 and the Wnet should be 3203J 😳😳😳😳
It depends on which side of the equation that term is placed. If you place the "energy lost" term on the left side it becomes negative and if you place it on the right side it becomes positive.
Love from india
Thank you and welcome to the channel!
Weer een Nederlander....:P
Eignelijk ik was geboren in Belgie.
Osm
🙂
Glad you liked the video.
@@MichelvanBiezen 😇
Did this and got the wrong answer lol
Than happens. Try again.
Who's here because of online class due to ncov
Your video too dark hard too watch it
Yes our older videos needed better lighting. We have fixed that problem since
Why using the equation "(F - friction) * distance" has different outcome? Isn't it about 3206? This is from that you plus "cos30*0.2" rather than minus them although the "cos30*0.2" is from the friction that is the opposite direction from the force. Could you explain it to me?
The force is pushing the block up the hill, in the same direction as the motion. The friction force pushes in the opposite direction of the motion of the block, therefore it does negative work on the block.
@@MichelvanBiezen Oh, the reason why you plus the heat loss rather than minus the heat loss is because we want to figure out how much amplitude we need to move the box. Am I right?