If the body is at rest and the force applied by us is equal to mg, the body will remain at rest. If the body is moving at a constant velocity, and the force applied by us equals the force of gravity (mg), then the object will continue to move at a constant velocity.
It depends on what work you are calculating: 1) If you are calculating the work done by the force lifting the object, then the direction of the force and the direction of the displacement is the same and the angle between them is zero degrees. 2) If you are calculating the work done by the force of gravity as the object is pushed up (by another force), then the direction of the force of gravity is opposite to the direction of motion and the angle between them is 180 degrees. Consequently the work done by the force of gravity will be negative.
Shane, In this example, just enough force is applied to move the object upward at a constant velocity. That means that there is no net force acting on the object and therefore there is no acceleration.
I thought we the applied force is exactly equal to the weight of the object; it means the net force is zero, and according to Newton 2rd law, if the net force is zero, then the object will either stay at rest or move in a straight with constant velocity!. In this case, I believe the object should stay at rest since the net force is zero! Could you please verify my confusion? thank you very much!
Once the object is moving upward (after a momentary extra force to get the object moving), the object will continue to move upward at a constant speed when the net force is zero.
8 ปีที่แล้ว +3
Sir if move a body vertically upwards starting with v=0, till point B if I calculate at a point M between A &B where v=10m/s so if the workdone is calculated at that point.won't the workdone be more than zero? because, Workdone=KE final - KE initial , W=10-0, W=10 isn't gravity conservative force
It is always good to mention what force is doing the work. In this example it is the force of gravity. We are also assuming that the object is moving up at a constant (slow) speed, so that there is no change in the KE.
8 ปีที่แล้ว
but the initial velocity is zero and the final has a value?
Brilliant as usual. Only confusion is that if gravity is negative y direction and therefore the force opposing gravity should be positive as it is pointing upwards? Anyways enjoying every moment of this? Moment in this case is a factor of time and happiness not mass.
how will the object be lifted if applied force is lesser than weight ? may i confirm that first the applied force is greater than weight and then applied force is decreased to make it equal to weight then the object moves with constant velocity upto desired height h and hence the work done is approximately equal to increase in potential energy mgh without increaing kinetic energy?
Potential energy is positive if an object is doing work against a conservative force, such as gravity in this case. So when height is positive and an object is lifted up, the potential energy will also be positive because it's work done against gravity. But the work done by gravity will be negative because the direction of motion and the force of gravity is opposite each other. So work done by gravity is kind of the opposite of mgh.
Since there is some velocity while moving up shouldn't there be K.E as well? When we reach 'h' height the velocity becomes zero so K.E becomes zero? Is this logic correct?
Good questions. In this example, it is assumed that the velocity at which the block is moved up is very small and insignificant and it is kept constant. Thus we ignore it.
If you are pulling something up against gravity, the person pulling is doing positive work (adding potential energy to the object), and gravity is doing negative work since gravity acts downward and the direction of motion of the object is upward.
Sir, i don't understand how the angle between gravitational energy (pointing in the opposite direction) with the force is 0. Shouldn't it be 180 ? Or in this case we should consider that all the work is done upward ?
Isn't this the same equation for PE = mgh? if yes whats their difference and May I just clarify if the object is far from the ground then the work done by the gravity is a negative while if it stays on the ground its positive?
If the object is motionless on the ground, gravity does NO work on it. If the object is motionless far from the ground, then gravity does NO work on it.
An object can use its potential energy to do work. When an object is lowered it losed potential energy PE = mgh (when h becomes smaller the PE drops). That loss of energy can be used to do work.
+mr orange I am not familiar with the "A" level, but there are many hundreds of videos in the playlist on E & M. PHYSICS - ELECTRICITY AND MAGNETISM, OPTICS, ATOMIC AND NUCLEAR PHYSICS, AND QUANTUM MECHANICS
I understood that we need to multiply by the cos of theta but i don't understand these unit vectors ŷ . ŷ being 1 . 1 times cos of theta :S Thank you for the channel anyway
Theta is the angle between the direction of the Force and the direction of the displacement. In this problem the angle is zero degrees and 180 degrees (for the work done by gravity). That is where the negative sign comes from.
Excuse me sir, I just want to know is G negative or positive like, for example, if g is negative but why the gravitational potential energy always be positive ?
When used in the equations of kinematics, g is considered negative (- 9.8 m/sec^2) When used in the PE equation, we use the magnitude of g and thus it is positive.
I have the same question as Khaled Sankar but I still don't understand the answer: If the direction of the Force of gravity is in the negative direction, but the displacement is in the positive direction, why isn't the cosine between the vectors 180?
So as I understand it, when you raise a off of the ground, you are adding energy to that system. If the work done by the force is equal and opposite to the work done by gravity, does that mean that the total work is zero? Wouldn't that mean there is no change in energy in the system?
You can only consider one force at a time. Work is done by the force pushing the box up. W = F x d The work done by gravity is negative work since the force of gravity acts downward and the box is moving upward. The magnitude of that negative work is the same as the magnitude of the work done by the force.
Am I right if I said the work done by gravity when we left objects upward is negative potential energy, and positive PE in case we lower the object down ?
When the force (that is doing work) and the displacement are in the same direction, then the work done is positive. Thus gravity does work, when the object is lowered.
Eric, I am not following your question, but let me try to answer it this way. When calculating the work done, work is the force required to lift up the object multiplied with the displacement. Since the force is upward (positive and acting against the force or gravity), the work done will be positive.
I don't understand 1 point. F sub G = -mgy and d=hy are in the opposite direction. Hence, in my opinion, why it isn't cos180 instead of cos0? Michel van Biezen
Khaled, Work done is the dot product of the force doing the work and the displacement vector. In this case the force doing the work is upward and the direction is upward.
To get the object moving you need a momentary force, once the object is moving, the object will continue to move (at a constant speed) with a zero net force. (Newton's first law).
Nameri, Very good question. If the object is not moving initially, then when the forces are equal, the object will not move. But it is better to think about it this way. If there is no net force then there will be no acceleration. But once the object is moving (because of a small momentary extra force), Newton's first law dictates that an object once in motion will remain in motion, unless a force changes that. With no net force the object will continue to move upward.
Michel van Biezen But didnt we use this theory in problems with an object moving with a constant speed ( not standing ). I mean the object is not moving initially in this problem as i see. I make a mistake somewhere but couldnt see. What is the small momentary extra force which initially moves the object in this question ?
nameri lafazan For the purpose of the example, we ignore this "initial small force" to get the block going. It can be infinitesmally small force for a very short period of time. So we pick up the problem with the object moving upward slowly at constant speed so no force is required to keep it going, and thus only force if required to overcome the force of gravity. This is usually not mentioned in the book, so it becomes confusing. The best way to look at it is to assume that the object is already moving upward at t=0.
Michel van Biezen I don't get what you're saying. The object continues to move upward only because you are continuing to apply an upward force -- otherwise, the force of gravity will immediately pull it back down!
Watch this videos, and the ones following this one. Physics - Mechanics: Vectors (12 of 21) Product Of Vectors: Dot Product th-cam.com/video/hho9a0_y7KY/w-d-xo.html
But if the work done by gravity is negative the work done by you, that means there is no net work done on the object, and if so, how can we say that the object has gained in potential energy when no net work has been done on it?
7beers The work done by the force pushing the block up = F * d = W = mgh = PE gained by the block. The block obviously gains potential energy since it moves up. It is the same as pushing a block horizontally with the force pushing being equal to the force of friction. W = F * d except in this case the work done by the force is converted to an equal amount of heat.
this video was uploaded 6 years ago now and I still see you commenting on recent peoples comments. Outstanding! These videos really help out.
These videos deserve so much more exposure. I wish you were my professor! Very clear explanations
If the body is at rest and the force applied by us is equal to mg, the body will remain at rest.
If the body is moving at a constant velocity, and the force applied by us equals the force of gravity (mg), then the object will continue to move at a constant velocity.
It depends on what work you are calculating: 1) If you are calculating the work done by the force lifting the object, then the direction of the force and the direction of the displacement is the same and the angle between them is zero degrees. 2) If you are calculating the work done by the force of gravity as the object is pushed up (by another force), then the direction of the force of gravity is opposite to the direction of motion and the angle between them is 180 degrees. Consequently the work done by the force of gravity will be negative.
+Michel van Biezen Thank you!
Shane,
In this example, just enough force is applied to move the object upward at a constant velocity. That means that there is no net force acting on the object and therefore there is no acceleration.
Can I use the Kinetic energy principle to calculate the work done?
I thought we the applied force is exactly equal to the weight of the object; it means the net force is zero, and according to Newton 2rd law, if the net force is zero, then the object will either stay at rest or move in a straight with constant velocity!. In this case, I believe the object should stay at rest since the net force is zero! Could you please verify my confusion? thank you very much!
Once the object is moving upward (after a momentary extra force to get the object moving), the object will continue to move upward at a constant speed when the net force is zero.
Sir if move a body vertically upwards starting with v=0, till point B if I calculate at a point M between A &B where v=10m/s so if the workdone is calculated at that point.won't the workdone be more than zero? because, Workdone=KE final - KE initial , W=10-0, W=10 isn't gravity conservative force
It is always good to mention what force is doing the work. In this example it is the force of gravity. We are also assuming that the object is moving up at a constant (slow) speed, so that there is no change in the KE.
but the initial velocity is zero and the final has a value?
Brilliant as usual. Only confusion is that if gravity is negative y direction and therefore the force opposing gravity should be positive as it is pointing upwards? Anyways enjoying every moment of this? Moment in this case is a factor of time and happiness not mass.
how will the object be lifted if applied force is lesser than weight ? may i confirm that first the applied force is greater than weight and then applied force is decreased to make it equal to weight then the object moves with constant velocity upto desired height h and hence the work done is approximately equal to increase in potential energy mgh without increaing kinetic energy?
Potential energy is positive if an object is doing work against a conservative force, such as gravity in this case. So when height is positive and an object is lifted up, the potential energy will also be positive because it's work done against gravity. But the work done by gravity will be negative because the direction of motion and the force of gravity is opposite each other. So work done by gravity is kind of the opposite of mgh.
Since there is some velocity while moving up shouldn't there be K.E as well?
When we reach 'h' height the velocity becomes zero so K.E becomes zero?
Is this logic correct?
Good questions. In this example, it is assumed that the velocity at which the block is moved up is very small and insignificant and it is kept constant. Thus we ignore it.
Thanks professor :)
If I pull up a chain freely hanging on a table edge, is the gravity doing positive work or negative work in that case ?
If you are pulling something up against gravity, the person pulling is doing positive work (adding potential energy to the object), and gravity is doing negative work since gravity acts downward and the direction of motion of the object is upward.
@@MichelvanBiezen THANKYOU SO MUCH! I'VE BEEN CONFUSED FOR SO LONG!
Yes, I remember being confused about that as a student as well.
Sir, i don't understand how the angle between gravitational energy (pointing in the opposite direction) with the force is 0. Shouldn't it be 180 ? Or in this case we should consider that all the work is done upward ?
Isn't this the same equation for PE = mgh? if yes whats their difference and May I just clarify if the object is far from the ground then the work done by the gravity is a negative while if it stays on the ground its positive?
If the object is motionless on the ground, gravity does NO work on it. If the object is motionless far from the ground, then gravity does NO work on it.
@@MichelvanBiezen WOW THIS MADE THINGS SIMPLER I'm Glad you are here! Thank you!
Glad to be of help,
you save my life! good video!
Glad I could help!
will the force of gravity and the force applied by us cancel each other resulting in the body not moving
I don't know if i can get a reply but how can someone calculate work done by weight of objects
An object can use its potential energy to do work. When an object is lowered it losed potential energy PE = mgh (when h becomes smaller the PE drops). That loss of energy can be used to do work.
Mr.Biezen , do you make videos on A level electricity e.g resistance , p.d etc ?
+mr orange
I am not familiar with the "A" level, but there are many hundreds of videos in the playlist on E & M.
PHYSICS - ELECTRICITY AND MAGNETISM, OPTICS, ATOMIC AND NUCLEAR PHYSICS, AND QUANTUM MECHANICS
I didn't understand Wfg how did you substitute mg y^ in place of Fg
We didn't "substitue", but calculated the work done, both by the Force pushing and the Force of gravity so you can compare them.
I understood that we need to multiply by the cos of theta but i don't understand these unit vectors ŷ . ŷ being 1 . 1 times cos of theta :S
Thank you for the channel anyway
Why is it W = Fdcos(theta). What angle is theta?
Theta is the angle between the direction of the Force and the direction of the displacement. In this problem the angle is zero degrees and 180 degrees (for the work done by gravity). That is where the negative sign comes from.
Excuse me sir, I just want to know is G negative or positive like, for example, if g is negative but why the gravitational potential energy always be positive ?
When used in the equations of kinematics, g is considered negative (- 9.8 m/sec^2) When used in the PE equation, we use the magnitude of g and thus it is positive.
So PE is scalar right ?
What about normal force sir? So the normal force is also using g as a negative quantity? (N=mg)
If you are calculating the magnitude of the force, yes.
Thank you sir
I have the same question as Khaled Sankar but I still don't understand the answer: If the direction of the Force of gravity is in the negative direction, but the displacement is in the positive direction, why isn't the cosine between the vectors 180?
So as I understand it, when you raise a off of the ground, you are adding energy to that system. If the work done by the force is equal and opposite to the work done by gravity, does that mean that the total work is zero? Wouldn't that mean there is no change in energy in the system?
You can only consider one force at a time. Work is done by the force pushing the box up. W = F x d The work done by gravity is negative work since the force of gravity acts downward and the box is moving upward. The magnitude of that negative work is the same as the magnitude of the work done by the force.
@@MichelvanBiezen Nice, that makes sense. Thanks so much.
Am I right if I said the work done by gravity when we left objects upward is negative potential energy, and positive PE in case we lower the object down ?
When the force (that is doing work) and the displacement are in the same direction, then the work done is positive. Thus gravity does work, when the object is lowered.
so when we moved up a box of an incline plane, why the gravity is +ve , not -ve ?
as the previous example
Eric,
I am not following your question, but let me try to answer it this way.
When calculating the work done, work is the force required to lift up the object multiplied with the displacement. Since the force is upward (positive and acting against the force or gravity), the work done will be positive.
ok i understand sir, tq very much
as initially the block is resting on the ground. wont there be any role of reactionary force due to mg?
Not if there is already another force of equal magnitude pushing upwards (as shown in the video)
I don't understand 1 point. F sub G = -mgy and d=hy are in the opposite direction. Hence, in my opinion, why it isn't cos180 instead of cos0? Michel van Biezen
Dan Huynh
It depends on which force is doing the work.
Which force is pushing the mass upward?
Michel van Biezen Ok I got it. Thank you for taking time answering my question. I love ilectureonline so much.
if the object is moving upward and the force of gravity is pointing downward how could the angle between them be 0, shouldn't it be 180 ?
Khaled,
Work done is the dot product of the force doing the work and the displacement vector.
In this case the force doing the work is upward and the direction is upward.
ohhhh, i got it thanks
i don´t finish understand why the angle between the gravity force and displacement is o and not 180?
How would the object accelerate when the net force is equal to zero !!
To get the object moving you need a momentary force, once the object is moving, the object will continue to move (at a constant speed) with a zero net force. (Newton's first law).
How come the force F = mg is acting in the positive y direction? shouldn't it be the opposite (as in weight)
The force required to lift the object against gravity must be positive (upward direction).
Very helpful. Thank you!
Glad it was helpful!
Thanks sir, i appreciated.
You're welcome. Glad you found the video helfpul. 🙂
if the work done by the force and the work done by the gravity is equal,how does the object move ?
Nameri,
Very good question.
If the object is not moving initially, then when the forces are equal, the object will not move.
But it is better to think about it this way. If there is no net force then there will be no acceleration. But once the object is moving (because of a small momentary extra force), Newton's first law dictates that an object once in motion will remain in motion, unless a force changes that. With no net force the object will continue to move upward.
Michel van Biezen But didnt we use this theory in problems with an object moving with a constant speed ( not standing ). I mean the object is not moving initially in this problem as i see. I make a mistake somewhere but couldnt see. What is the small momentary extra force which initially moves the object in this question ?
Michel van Biezen i saw the same thing in another video of you,i guess i dont know what is Fnet.Can you explain ? :) thanks for sparing your time btw
nameri lafazan
For the purpose of the example, we ignore this "initial small force" to get the block going. It can be infinitesmally small force for a very short period of time.
So we pick up the problem with the object moving upward slowly at constant speed so no force is required to keep it going, and thus only force if required to overcome the force of gravity.
This is usually not mentioned in the book, so it becomes confusing.
The best way to look at it is to assume that the object is already moving upward at t=0.
Michel van Biezen I don't get what you're saying. The object continues to move upward only because you are continuing to apply an upward force -- otherwise, the force of gravity will immediately pull it back down!
Merci monsieur
C'est mon plaisir.
Isn't the work done by gravity the same as potential energy?
It is the opposite. The force working against gravity adds potential energy to the object. (not gravity)
thank you this helped me!
i cant understand the dot dot method..
Watch this videos, and the ones following this one. Physics - Mechanics: Vectors (12 of 21) Product Of Vectors: Dot Product th-cam.com/video/hho9a0_y7KY/w-d-xo.html
But if the work done by gravity is negative the work done by you, that means there is no net work done on the object, and if so, how can we say that the object has gained in potential energy when no net work has been done on it?
7beers
The work done by the force pushing the block up = F * d = W = mgh = PE gained by the block.
The block obviously gains potential energy since it moves up.
It is the same as pushing a block horizontally with the force pushing being equal to the force of friction.
W = F * d except in this case the work done by the force is converted to an equal amount of heat.
Thxxxxx😍😍♥️♥️♥️♥️♥️
Glad it was helpful! 🙂
I would love to be your student...
u look like my yoga instructor
I wish I was in the same shape as your yoga instructor
✔
🙂
This example would have been better with numbers.
There are lots of other examples (with numbers).