Why are we adding Friction here ( 18:52 ) ? Considering Friction is against the Force shouldn`t the work done by friction equal to negative work? I can understand that we need more work (energy) considering there is a force acting against the motion intended by intuition but I can`t comprehend it with physics. From what you said in your work & energy video Friction is negative because its 180´, in opposite direction to F. Therefore its negative work, it eats up energy.
This is sometimes taught using the work-energy theorem which states that the net work done will equal the change in KE. The net work done would be given by the net force x the distance moved, so we have that, Push x distance + comp. of gravity x distance + friction x distance = 0 since there is no change in KE. The component of gravity and the friction act down the ramp so their contributions would be negative. My more intuitive approach just says the work pushing up the ramp goes into heating by friction and a gain in GPE. I am using magnitudes. I am saying the size of this must equal the size of this + this.
Why did we add the original force of mg(3/6) to 10 wouldn't we substract 10 as it is the resultant force? And then we would end up with a resultant of 0? How should i see this?
A 64 kg student travels from the first floor to the fourth floor of a school (a height of 15 m). What total work did she do climbing the stairs? Is this answer to this 9600J? Thank you
@@queuethemusic The work is the work done by the person, and if they lift the weight at constant speed, the force that they will pull with is the tension.
Hello sir I don’t understand why when we push until the resultant force is 0 the object would move? Wouldn’t we just be supporting the object’s weight rather than accelerating it from rest, which would require a resultant force bigger than 0?
We don't need to accelerate it just have it move at constant velocity which requires zero net force. We ignore energy to get it going since this can be made infinitely small.
Keep in mind there is no net force, so we are really talking about two equal forces. One is friction and the other is the pushing force which will be in the direction of motion.
18:22 Mr. Doner why did you add 10N of Frictional force, shouldn't it be -10N? as it is opposing the push force?
There are just the two forces and they are both down the incline. The component of gravity acts down the incline as does the frictional force.
Thank u very much, sir. Our school wouldn't ever have such a magnificent lecture.
You are very welcome!
Why are we adding Friction here ( 18:52 ) ? Considering Friction is against the Force shouldn`t the work done by friction equal to negative work? I can understand that we need more work (energy) considering there is a force acting against the motion intended by intuition but I can`t comprehend it with physics. From what you said in your work & energy video Friction is negative because its 180´, in opposite direction to F. Therefore its negative work, it eats up energy.
This is sometimes taught using the work-energy theorem which states that the net work done will equal the change in KE. The net work done would be given by the net force x the distance moved, so we have that, Push x distance + comp. of gravity x distance + friction x distance = 0 since there is no change in KE. The component of gravity and the friction act down the ramp so their contributions would be negative. My more intuitive approach just says the work pushing up the ramp goes into heating by friction and a gain in GPE. I am using magnitudes. I am saying the size of this must equal the size of this + this.
Why did we add the original force of mg(3/6) to 10 wouldn't we substract 10 as it is the resultant force? And then we would end up with a resultant of 0? How should i see this?
Ohh just fyi this is in minute 18:35
We need to push harder because of friction so we do more work. We need to push with a force of 20 N over a distance of 6m.
A 64 kg student travels from the first floor to the fourth floor of a school (a height of 15 m).
What total work did she do climbing the stairs?
Is this answer to this 9600J?
Thank you
Yes, that would be the work done against gravity.
For the pulley problem, do we also need to include the force of the weight of the object as well as the tension for total work?
at what time in the video?
Chris Doner at 19:43
@@queuethemusic The work is the work done by the person, and if they lift the weight at constant speed, the force that they will pull with is the tension.
Hello sir I don’t understand why when we push until the resultant force is 0 the object would move? Wouldn’t we just be supporting the object’s weight rather than accelerating it from rest, which would require a resultant force bigger than 0?
We don't need to accelerate it just have it move at constant velocity which requires zero net force. We ignore energy to get it going since this can be made infinitely small.
@@donerphysics ahh that makes sense. Thank you!
For the "Power at constant velocity" does the F and v have to be in the same direction?
Keep in mind there is no net force, so we are really talking about two equal forces. One is friction and the other is the pushing force which will be in the direction of motion.
14:12 that was sneaky af
Very sneaky.
so in the end what would the efficiency of the massless frictionless pulley be?
100%
At 15:20 you say mgsin theta then you say mg 3/6, shouldn't it be mg sin 3/6??
No, sin theta is the ratio of opposite to hypotenuse.
@@donerphysics thanks I got it now
why is it mg sin theta and not mg cos theta?
At what time are you referring to?
@@donerphysics 15:24 , when talking about efficiency on an incline
The angle of the incline is the same as the angle that is opposite to the vector component along the incline's surface.
@@donerphysics But isn't work defined as F*d*cos theta
If the velocity is constant, the net force would be zero. Would the power still exist?
Power is needed to overcome friction. F in the equation is the push force, or equivalently the frictional force, not the net force.
Then would it be power output or power input? Thank you!
Output.
Which grade is this
American syllabus?
You should not use videos of monkeys or any animals being forced to do anything like pushups that is unethical tbh
He signed a consent form.
@@donerphysics I don't think the monkey signed a consent form but anyway your videos are very helpful so thank you
I want to see them bench press