Work done by static friction can be zero,-ve, or +ve. Imagine a two block system one lying over other and a force less than the magnitude of static friction is applied on one block in this case it is not zero. However work done by static friction on a whole system is zero.
A skier starts from rest at the top of friction-less incline of height 20 m. at the bottom of the incline, she encounters a horizontal surface. Show in figure How far does she travel on the horizontal surface coming to rest
جعفر جبر الخزعلي, Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
Excellent! Yes, we didn't consider that in this problem, but somebody had to put work into it to get it moving at vinitial. They had to do work of at least 1/2 mv^2, more if there was friction. Cheers, Dr. A
It can't be zero because you have to work against friction. If you slide your book across the table and back, you have certainly done work on it. The work that you have done ultimately goes into heat due to friction: the book and the table both heat up. Cheers, Dr. A
The kinetic frictional force is always fk = μk * N where μk is the kinetic frictional coefficient and N is the normal force. If you're on a level surface, then N=mg. The cos 180 means that the force is opposite the direction of travel, since friction works against you. Cheers, Dr. A
Can you help me with this sir? a 6,000-kg car is moving at 10m/s. when the driver steps on the brakes, the car continues to move a distance of 1.2 m until it eventually stops. what is the amount of friction that the road exerted on the car?
10 months late, I am only a Junior in highschool (g11). First find the Kinetic Energy (mv^2/2) = 6000x10^2/2) = 300000J. Wnet = Fd, Wnet = Sum of Energy, since on flat ground there is only Gravitational energy. So 300000J = F(1.2m) Ff = -250000N since friciton is opposite direction of displacement. This may or may not be right hope it helps even if it is too late
Sugandha, It depends on which side of the equation you put it. Remember, energy is always conserved, just keep track of it. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
if you are asking when the kinetic friction will be positive, you can consider a situation where a body is resting on top of another body, then if the bottom body is moved forward and the top body slides back, then the frictional force on the top body acts along the direction of the motion of the first body and we can say that the net friction work is positive for the top body.
Yes, when you walk you exert a force away from you and onto the floor and the friction against the floor and your foot allows you to move forward. And just to get ahead of the people you are going to question whether or not this is actually friction and not say normal force, note how difficult it is to walk on a surface with a low coefficient of friction like ice.
I have been stuck on a problem for the past hour and after only watching HALF of your video I was able to solve it. THANK YOU!!
Best physics teacher ever ! Thank you so much :))
please continue to make more videos!! you are the most helpful physics professor on youtube!!
Now that makes me smile, and also a little concerned. Hopefully more talented teachers will get online.
Cheers,
Dr. A
Thank you so much for doing this videos. Really appreciated!
Ale M. Preve
I read in a book that work done by static friction can be positive negative as well as zero is it incorrect please answer th
sir I am INDIAN and this is very helpful for me in LOCKDOWN is also u hit like
no i wont hit like
Would the work done by kinetic friction also equal the change in kinetic energy, so in this case -0.5mv^2?
Work done by static friction can be zero,-ve, or +ve.
Imagine a two block system one lying over other and a force less than the magnitude of static friction is applied on one block in this case it is not zero.
However work done by static friction on a whole system is zero.
How did you teach yourself to write mirror-style so well?
He is just mirroring the video :D
In static friction, there is no motion, i.e ∆x = zero
i dont get how he's writing on the board and it is not the opposite. I'm confused.
What is delta x is it distance
A skier starts from rest at the top of
friction-less incline of height 20 m.
at the bottom of the incline, she encounters
a horizontal surface. Show in figure
How far does she travel on the horizontal
surface coming to rest
جعفر جبر الخزعلي,
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
what about the work you put into an object up to the point where it starts moving on a rough plane?
Excellent! Yes, we didn't consider that in this problem, but somebody had to put work into it to get it moving at vinitial. They had to do work of at least 1/2 mv^2, more if there was friction.
Cheers,
Dr. A
Can you do a problem on energy lost to friction
LEGEND.
If we move the same block back to the initial position then why will the total work done by friction force in the round trip be -ve and not 0
It can't be zero because you have to work against friction. If you slide your book across the table and back, you have certainly done work on it. The work that you have done ultimately goes into heat due to friction: the book and the table both heat up.
Cheers,
Dr. A
Im having trouble figuring out how to get the frictional force given delta x and the kinetic friction coefficient. and cos 180
The kinetic frictional force is always fk = μk * N where μk is the kinetic frictional coefficient and N is the normal force. If you're on a level surface, then N=mg. The cos 180 means that the force is opposite the direction of travel, since friction works against you.
Cheers,
Dr. A
How does he write on the screen like that? Lol wouldn’t it be reversed?? I can’t figure it out 🤣
Can you help me with this sir?
a 6,000-kg car is moving at 10m/s. when the driver steps on the brakes, the car continues to move a distance of 1.2 m until it eventually stops. what is the amount of friction that the road exerted on the car?
10 months late, I am only a Junior in highschool (g11). First find the Kinetic Energy (mv^2/2) = 6000x10^2/2) = 300000J. Wnet = Fd, Wnet = Sum of Energy, since on flat ground there is only Gravitational energy. So 300000J = F(1.2m) Ff = -250000N since friciton is opposite direction of displacement. This may or may not be right hope it helps even if it is too late
Static friction can do work, when a box is kept on another box
Work done by kinetic friction is negative it's ok but when we put it positive?
Sugandha,
It depends on which side of the equation you put it. Remember, energy is always conserved, just keep track of it.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
if you are asking when the kinetic friction will be positive, you can consider a situation where a body is resting on top of another body, then if the bottom body is moved forward and the top body slides back, then the frictional force on the top body acts along the direction of the motion of the first body and we can say that the net friction work is positive for the top body.
You can't meassure work done by friction like this
Sir does positive friction exist? Example?
Yes, when you walk you exert a force away from you and onto the floor and the friction against the floor and your foot allows you to move forward. And just to get ahead of the people you are going to question whether or not this is actually friction and not say normal force, note how difficult it is to walk on a surface with a low coefficient of friction like ice.
Good😛😛😛