Introductory Work due to Friction equals Change in Mechanical Energy Problem

[@efetugan8050]

I like your videos very much. The best lesson channel I've ever seen.

[@FlippingPhysics]

Thanks! 😃

[@obi-wankenobi7324]

++++ Obi Wan agrees.

[@ozansafoglu3741]

++++ OzanS agrees.

[@efetugan8050]

@@obi-wankenobi7324 may the force be with you

[@efetugan8050]

@@ozansafoglu3741 together, we are stronger.

[@renanaraujom]

What a great job, man! Keep it up! I really enjoy your videos. Thanks!

[@matthewwilson4592]

The best presentation on the subject I've ever seen. So thankful for people like you.

[@FlippingPhysics]

You are welcome. Thank you for your kind words.

[@oyska_g2985]

That pen was a nice touch. Very slick

[@nipungorantla5760]

I felt it when Bobby started freaking out over the answer at 8:12 😂

[@fredd298]

I had to pause at 7:14 to make sense of an equation. Upon looking at your paused faces in the excitement of a mass party, I felt like bobby, bo and billy where all three unique people with their own reactions to the situation. Nice acting.

[@FlippingPhysics]

Thanks! I do my best.

[@mustafaRoya]

Thank you mr p for making all those videos You really explained very well any one can understand that

[@FlippingPhysics]

+Gholam Mustafa Ali Thanks. 😀

[@Ndiedddd]

"Does anyone know where my pencil is?", That part got me laughing 😂

[@FlippingPhysics]

❤️

[@pawankumar-wt4fn]

Can you please explain how the work done by applied force is zero

[@FlippingPhysics]

Because, just like I explain in this problem, www.flippingphysics.com/tire-friction.html , there is no force applied acting on the hockey puck.

[@sachinprabu05]

When a ball is thrown upwards with an velocity and moves certain height and then velocity of a ball is zero.So Total work done by the ball is negative or positive? I Think negative,because of force of gravity acting downwards through out the motion. So the angle between the the d and F is 180°. Anyway Is this answer is correct? Reply pls...

[@carultch]

When a ball moves upward, the ball is doing work on the gravitational field. Its kinetic energy is being converted into its gravitational potential energy. The work done by the ball is positive, because energy is leaving its kinetic form and converting into its potential form. The reason why you are getting a negative sign, when calculating W=F dot d, is that this equation refers to work done ON the object as positive, rather than work done BY the object. Work done BY the object is what you get when the work done on the object is negative.

[@AlexmaryGR]

Thank you very much! I love your videos!

[@FlippingPhysics]

You are absolutely welcome. Glad to help you learn!

[@Uncertaintycat]

ok so it's driving me nuts and I need to know. What is Bo staring at all this time??

[@FlippingPhysics]

The clock.

[@kevinwilliam77]

Can anyone explain why did he divided both sides by -m, I don't get how he could do that

[@FlippingPhysics]

+Kevin William Look at the equation on the screen at 7:18. On the left hand side of the equation in the numerator we have mass times negative one, which is negative mass. On the right hand side of the equation in the numerator, we have negative mass. Therefore, we can divide the whole equation by negative mass, which cancels out the mass and the negative from both sides of the equation.

[@sivakumarsiva1743]

Is it noving witha constant velocity

[@FlippingPhysics]

The puck is not moving with a constant velocity. It is moving with a constant acceleration.

[@feyzayucel7139]

Hi Mr.P I have a question, when do you think you'll be going over rotational motion, simple harmonic motion, and waves? Normally, your content and my class go at around the same pace, so it's very helpful, but we've already covered rotational motion (kinematics, dynamics, torque, etc..) and simple harmonic motion, and are almost done with waves. Will you be covering everything before the AP exam, or do you plan on explaining other concepts soon? Because your reviews really help. Thank you:)

[@FlippingPhysics]

+Feyza Yucel Previously I have created review videos of every AP Physics 1 topic. flippingphysics.com/ap-physics-1-review.html Now I am slowly working my way through detailed videos of every AP Physics 1 topic, however, it takes me a long time. flippingphysics.com/making-a-video.html I'm doing my best, but it's a slow process. Glad you find my videos helpful for learning.

[@bvolpato]

Great video, thanks!

[@FlippingPhysics]

You are welcome!

[@sivakumarsiva1743]

Why is there no work done by force applied clearly you are applying a force on the puck and it moves through some displacement so i think there must be a force

[@FlippingPhysics]

The duration of the problem is from the moment I let go of the puck until it stops moving. Therefore, no force is applied to the puck during the problem. The force is applied to the puck before the problem begins in order to give the puck an initial velocity.

[@zhgshdbssss6536]

I solved it using the work energy theorem and got the same answer. W net = (1/2)m(v^2 - u^2) and the W(friction) = F(kinetic friction)*s*cos theta . The force of gravity and the force normal do not do any work on the object as both of these forces act perpendicular to the displacement. Simplifying for the work done by friction we get, W(friction) = mu*normalforce*s*-1 ( since cos 180 = -1) but Normal force = mg since there is no acceleration in the ‘Y’ direction. Therefore; - mu*m*g*s = (1/2)m(v^2 - u^2) => (v^2 - u^2) = - 5.12*m = - mu*m*g*s and, everybody brought mass to the party! => s = 5.12/mu*g = 5.12/0.6*9.8 = 0.87 metres and the displacement = - 0.87 metres.

[@FlippingPhysics]

Very nice! This video precede the Net Work equals Change in Kinetic Energy equation, in my curriculum, however, you can certainly solve the problem that way.

[@zhgshdbssss6536]

Flipping Physics Thank You sir! But I have a doubt: I was reading the derivation of the work energy theorem in my tb and it goes like this W = Fs => W(net) = mas Since v^2 - u^2 = 2as => as = (v^2 - u^2)/2 Substituting in the original equation, we get: W(net) = (mv^2)/2 - (mu^2)/2 Which is delta K.E. But where did the cos(theta) go which is there in the original equation of the work?? Shouldn’t the derivation be : W = F*s*cos(theta) And then we do the same as before and we should get: Net work = (1/2)m(v^2 - u^2)*cos(theta) , where theta is the angle between the NET force and the displacement, right? What is wrong with my reasoning? I know this probably isn’t correct because let’s take a simple case: v = 0 and u is a non-zero positive integer. Let’s say theta = 180 degrees,i.e., according to my derivation, we would expect the work done to be negative, since cos 180 = -1. But when you do v^2 - u^2, you would get a negative value itself which would then give us a positive net work which is incorrect. I think part of the wrong reasoning is that whenever we use the work equation, we only plug in the MAGNITUDES of the force and displacement as the positive or negative work is only decided by the term cos(theta). But then how do I get v^2 - u^2 as negative? If u is let’s say 10m/s then v^2 - u^2 would be -100. I can’t just write +100 for that. Could you please clarify my doubt sir? This doubt came in my mind from this particular question only. When I refer to ‘u’ , I mean initial velocity btw and v is final velocity. I know my above answer to this question that -5.12m = -mu*mg*s is correct and if i did net work = (1/2)m(v^2 - u^2)*cos(theta) I would get : net work = -5.12*mass x (-1) since theta = 180 degrees because there is no force applied(the puck has inertia of motion) so net force in the x direction equals force of friction = 5.12*m. And then if I had equated this with -mu*mg*s I would have gotten a negative DISTANCE( since we only use the magnitudes in the work equation) how????

[@FlippingPhysics]

@@zhgshdbssss6536 Dude. I am sorry, however, I do not have the time to be able to parse through all of that. If I took the time to do so whenever people asked, I would not have time to make videos. Please watch my video where I derive the Work Net = Change in Kinetic Energy equation. I hope it helps: www.flippingphysics.com/wnet-ke.html

[@Dap-ri2yh]

0:18 Bo u really gonna take that?

[@stealth_chain]

Did the students coordinate with each other to come to class with the same pair of mismatched socks? 🤔

[@FlippingPhysics]

Pure happenstance.

[@Ipfreely31415]

Don't you mean Bo got his hairs cut? Get it? No? Ok

[@carultch]

I thought it might be foreshadowing in to what the concept of negative work would mean. Work done on the object vs work done by the object. A haircut done on Bo, or a haircut done by Bo.

[@TheeSmokeWilddxx]

Calm down Bobby jeez.