I find this video extremely valuable. So many resources I’ve seen so far only state that “when a man walks with a box in his hand, work done on the box is zero”. This statement is incomplete. It should state “when a man walks with a constant velocity carrying a box in his hand, work done on the box is zero”. You have demonstrated this so well in this video. Thank you for your effort. Best regards.
Here's what I remembered before watching the vid. If the component of the direction of motion and the component of the direction of force is perpendicular to each other, no work is done. In order to do work, an object must displace(move) in the same direction as that of the force acting on it. Your arms get tired because your muscles are "doing work against each other."
Since I don't know the circumstances, there is not enough information to comment. You appear to indicated that the speed and the height doesn't change which means there was not change in energy.
Here's a question that I have been pondering for a while on this problem. If there is no Work done, then how come we pay people to bring boxes of items to our homes? Shouldn't it be free since in this example they are not doing work?
PH3A: Could we also use the equation F=ma to determine that the force is zero? Since velocity is constant, acceleration would be zero. So F=(m)(0) = 0 which would make W = 0.
@@MichelvanBiezen but there are no horizontal forces acting on the box itself, but only the person. Would the person and the object be one system as a whole? I can see why work is being done on the PERSON, when a≠0. But the object itself only has forces Fg/Fn which are vertical components, and the box is not moving in those directions. Please tell me what I’m missing
there is work done when you Weight(f)* lifting(h) >>> w=f*h with your Definition if no change in place so there is no work done ,,can you explain this ???
It all depends on when "you start the clock". Yes, when you lift an object against gravity you are doing work. But when you hold an object at the same height and just walking around with it in a straight line at the same speed, you are not doing work.
You say at 0:46 that according to 3rd law, Normal reaction=mg. But actually Normal reaction isn't a reaction force. Is it? Anyways, your videos are so awesome. :D Thank you
Yes Saurabh! Actually I agree with your point :) Inclined plane situation is a good counter example to explain normal force is not the reaction force of weight. Another example is that in the situation of taking lift and is in acceleration, normal force is not equal to weight too~ Actually, other that comparing the values between forces to determine whether they are action and reaction force pair, we can also check if the force pair: 1. are acting in exactly opposite direction 2. are acting on two different(interacting) objects. 3. exist or disappear at the same time. In inclined plane situation, point 1 is violated. Both normal force and weight are acting on the same object, point 2 is violated in any situation. In case of free-falling, weight exists but no normal force (Assume no air resistance). point 3 is violated. All the above examples explain why normal force and weight are not action-and-reaction force pair. Anyways, thanks Michel so much for making these awesome videos~
Sir, but isn't it the case that when you start walking forward, yourbox ith applies a force to the box to keep the box moving with you. So in a sense, the frictional force between your hand and the box is doing work against the box's inertia
No, once the box is moving there does not need to be any friction force between your hand and the box. Once the box is in motion it requires no force at all to keep it moving in the same direction at the same speed. (That is Newton's first law).
Sir but there is some friction between feet and road hence there is a force of friction also and total work done is work done by frictional Force + gravitational Force and not zero
As I commented before just to move your whole body from point a to b there is work being done. His video is talking about a box sitting on a surface that is moving so the box is part of the system itself and it’s not being pushed forward
So work is done during starting with the box and I would assume stopping with the box? But not at constant velocity. If work is done during those times of starting ans stopping, is there also KE? Also, when the box is lifted, is KE also gained during the lift off the ground? Thanks!
Great, that makes sense. How does the energy conversion work though? If the box already has PE from its stored position relative to ground, and you stop and by doing so do work, causing the box to have KE, does this affect the PE at all?
If the person doesn't speed up there will not be an increase in kinetic energy, and there is no increase in potential energy because the height of the block doesn't change, therefore no work is done while walking at a constant velocity.
If Mary walks at constant velocity, the answer is: "no". Once the books are moving at constant velocity, it requires no force to keep them moving at that speed in the same direction (Newton's first law) and the direction of the force on the books by the head (vertically upward) is perpendicular to the direction of motion. W = F x d x cos (90)
if the normal force and force of gravity can cancel each other, then all gravity force should be cancel in every question bcz there r always got oppo of normal force...there will be no PE ....sorry i cannot get ,anymore can explain....thank you
The object has PE, but that was not the question. The problem is asking for word done moving the object from a to b. Since h does NOT change there is no difference in the PE between a and b.
On your second question the expression "gravity cancelling out" is not the way to express it. (We are not cancelling out gravity, since gravity always exists). Instead , we have to ask: "What is the NET FORCE on an object?" If the net force is zero (like for an object sitting on the ground), then there will be "NO ACCELERATION).
+Jonasa Karlsson Correct. A somewhat simplified way to think of it is that if something seems to be completely still relative to you, then you don't experience any acceleration. If you ride an airplane and you feel like you are sitting still motionless inside it, then the acceleration is almost zero, although you can still feel turbulence and turns to the left and right (or up and down), and those are also forms of accelerations.
Newton's first law states that an object in motion will remain in motion unless a force acts upon it to stop it. A car experience friction and wind resistance, and therefore will come to a halt.
Here's what I remembered before watching the vid. If the component of the direction of motion and the component of the direction of force is perpendicular to each other, no work is done. In order to do work, an object must displace(move) in the same direction as that of the force acting on it. Your arms get tired because your muscles are "doing work against each other." I don't think the problem is whether you're static or have made displacements. It does no work because of the above reasons. Sorry if I'm wrong; that's what I remembered from my textbook.
ok, I don't want to be a dickhead, but wanted to point this out. Obviously the box isn't doing any work, but you should probably look at the work the HANDS and therefore arms and legs are doing to know the work the person is doing, right? I mean obviously there's no work on the box.
It depends what you define as work and what the object is receiving the work. There is no work done on the box (no energy is gained by the box). But even standing there holding the object your muscles will get tired. But holding something up against gravity is strictly speaking not doing work (as defined in physics)
I find this video extremely valuable. So many resources I’ve seen so far only state that “when a man walks with a box in his hand, work done on the box is zero”. This statement is incomplete. It should state “when a man walks with a constant velocity carrying a box in his hand, work done on the box is zero”. You have demonstrated this so well in this video. Thank you for your effort. Best regards.
Glad you liked it.
Thumbs up... Truly helpful
Thank you so much. My professor just glazed past this idea without explaining why lol. This helped out a lot thanks
Glad I could help!
Here's what I remembered before watching the vid. If the component of the direction of motion and the component of the direction of force is perpendicular to each other, no work is done. In order to do work, an object must displace(move) in the same direction as that of the force acting on it. Your arms get tired because your muscles are "doing work against each other."
That is correct.
@@MichelvanBiezen Oh alright thanks!
these vids are actually soo helpful
Is this correct. Just to prove conservation of energy still works in this case.
Wd + Ea = Eb
Wd + .5mV^2 + mgh = .5mV^2 + mgh
Therefore Wd = 0
Since I don't know the circumstances, there is not enough information to comment. You appear to indicated that the speed and the height doesn't change which means there was not change in energy.
@@MichelvanBiezen Yes indeed I am assuming velocity and height does not change, therefore no work is done?
Here's a question that I have been pondering for a while on this problem. If there is no Work done, then how come we pay people to bring boxes of items to our homes? Shouldn't it be free since in this example they are not doing work?
That is off course the case with the strict definition of work. The person's arms would still get tired and appear to be doing work.
PH3A: Could we also use the equation F=ma to determine that the force is zero? Since velocity is constant, acceleration would be zero. So F=(m)(0) = 0 which would make W = 0.
Exactly. That is the key to understanding this concept.
This helped me a lot.. thank you so much
you specified that the velocity was constant, but if it were not, the work done would still be zero right, why is that?
If the velicity was not constant, then work would be done in order to accelerate the object, since the object would be gaining kinetic energy.
@@MichelvanBiezen but there are no horizontal forces acting on the box itself, but only the person. Would the person and the object be one system as a whole? I can see why work is being done on the PERSON, when a≠0. But the object itself only has forces Fg/Fn which are vertical components, and the box is not moving in those directions. Please tell me what I’m missing
When the acceleration is not zero, the there IS a horizontal force acting on the box. (F = ma) always works.
there is work done when you Weight(f)* lifting(h) >>> w=f*h
with your Definition if no change in place so there is no work done ,,can you explain this ???
It all depends on when "you start the clock". Yes, when you lift an object against gravity you are doing work. But when you hold an object at the same height and just walking around with it in a straight line at the same speed, you are not doing work.
@@MichelvanBiezen Thank you sir ..... you saved my life and believe in physics ...❤️❤️ Love from India ❤️
physics3a-why is there a normal force on the box? I thought normal force is only for when an object is on a surface?
In this case, the surface is the hand holding the box.
You say at 0:46 that according to 3rd law, Normal reaction=mg. But actually Normal reaction isn't a reaction force. Is it? Anyways, your videos are so awesome. :D Thank you
?
Though the magnitude of normal reaction is equal to mg, normal reaction itself is not a reaction force of the weight.
Mic, you're right, but not always, for example not on an incline
Yes Saurabh! Actually I agree with your point :)
Inclined plane situation is a good counter example to explain normal force is not the reaction force of weight. Another example is that in the situation of taking lift and is in acceleration, normal force is not equal to weight too~
Actually, other that comparing the values between forces to determine whether they are action and reaction force pair, we can also check if the force pair:
1. are acting in exactly opposite direction
2. are acting on two different(interacting) objects.
3. exist or disappear at the same time.
In inclined plane situation, point 1 is violated.
Both normal force and weight are acting on the same object, point 2 is violated in any situation.
In case of free-falling, weight exists but no normal force (Assume no air resistance). point 3 is violated.
All the above examples explain why normal force and weight are not action-and-reaction force pair.
Anyways, thanks Michel so much for making these awesome videos~
Saurabh Singh but there is a contact Force also between the hand and box
Sir, when the box is held by hand ,and not moving just stand, are there any work applied on object or zero?
The work would be zero.
Michel van Biezen well thank you
Sir, but isn't it the case that when you start walking forward, yourbox ith applies a force to the box to keep the box moving with you. So in a sense, the frictional force between your hand and the box is doing work against the box's inertia
No, once the box is moving there does not need to be any friction force between your hand and the box. Once the box is in motion it requires no force at all to keep it moving in the same direction at the same speed. (That is Newton's first law).
Sir but there is some friction between feet and road hence there is a force of friction also and total work done is work done by frictional Force + gravitational Force and not zero
Only the force pushing the object forward is considered. (Not the force between the feet and the road). (You made a good observation).
As I commented before just to move your whole body from point a to b there is work being done. His video is talking about a box sitting on a surface that is moving so the box is part of the system itself and it’s not being pushed forward
He told how much work is done to move the object which is in this hand not the work done by the body
So work is done during starting with the box and I would assume stopping with the box? But not at constant velocity. If work is done during those times of starting ans stopping, is there also KE? Also, when the box is lifted, is KE also gained during the lift off the ground? Thanks!
Yes, and also some KE would be gained by starting the box to move and that same amount of KE would be lost when you stop the box.
Great, that makes sense. How does the energy conversion work though? If the box already has PE from its stored position relative to ground, and you stop and by doing so do work, causing the box to have KE, does this affect the PE at all?
Also, would the KE be lost at constant velocity, or just constant ? Sorry for the many question. You just explain things so well!
Only if the force causes the box to change its height.
If the velocity remains constant (in any direction) the KE will remain constant as well.
Work is done and which is +ve .
If you ask how then I am ready to ans
But sir... What about the kinetic energy... And what will be the work done by the boy on the block...
If the person doesn't speed up there will not be an increase in kinetic energy, and there is no increase in potential energy because the height of the block doesn't change, therefore no work is done while walking at a constant velocity.
Michel van Biezen thank you sirrr.. Ur really awesome 😘😘
So according to this video, is it true that the same amount of effort is required to stand holding a box as to walk and carry it?
7beers
From a pure physics perspective yes.
(In real life off course not)
Isn't force required to keep the box moving in a horizontal direction because of air resistance?
+The Amazing Chemistry Show Wind resistance is ignored in these examples.
Why don’t we count in the force he exerted to start moving?
Yes, if you want to count the force required to start moving, then work is indeed done. But we assume the person is already moving.
What if the vertical forces weren't equal? For example a plane doesn't move horizontally so how do you calculate the work done?
Robert,
Take a look at some of the other videos like 15 and 16 and see if that answers your question.
Remember that W = F * d
or
W - delta (energy)
thats why you use trig, and get the forces x and y component. Fsin theta or Fcos theta
If Mary now walks across the room with the books on her head, does she do work? Why?
hope i can get a reply
If Mary walks at constant velocity, the answer is: "no". Once the books are moving at constant velocity, it requires no force to keep them moving at that speed in the same direction (Newton's first law) and the direction of the force on the books by the head (vertically upward) is perpendicular to the direction of motion. W = F x d x cos (90)
@@MichelvanBiezen thank youuu !
How abt the (PE)mgh? it got weight it got 9.8,it got also height...why dont have energy....>
if the normal force and force of gravity can cancel each other, then all gravity force should be cancel in every question bcz there r always got oppo of normal force...there will be no PE ....sorry i cannot get ,anymore can explain....thank you
The object has PE, but that was not the question. The problem is asking for word done moving the object from a to b. Since h does NOT change there is no difference in the PE between a and b.
On your second question the expression "gravity cancelling out" is not the way to express it. (We are not cancelling out gravity, since gravity always exists). Instead , we have to ask: "What is the NET FORCE on an object?" If the net force is zero (like for an object sitting on the ground), then there will be "NO ACCELERATION).
Sun filling can i get ur fb id
Your muscles are doing work and that requires energy :)
But we are not adding energy to the object and it requires no force to move the object forward at a constant speed.
how person g without doing force ??????
Sorry, didn't quite understand your question.
Thankyou so much
So if i was standing still why isnt there no work done to hold a box still?
+Jonasa Karlsson The definition of work is Force x Displacement. If the displacement is zero, then there cannot be any work done.
+Michel van Biezen But since also, there is no acceleration in either x or y direction, their is no work either right?
+Jonasa Karlsson
Correct.
A somewhat simplified way to think of it is that if something seems to be completely still relative to you, then you don't experience any acceleration.
If you ride an airplane and you feel like you are sitting still motionless inside it, then the acceleration is almost zero, although you can still feel turbulence and turns to the left and right (or up and down), and those are also forms of accelerations.
Thank you so much for your videos
if i ride car and don't put petrol it will stop so all time there are force prevent car from stopping and continue in constant velocity !!!!!!
Newton's first law states that an object in motion will remain in motion unless a force acts upon it to stop it. A car experience friction and wind resistance, and therefore will come to a halt.
tnx alor sr. 4m srilanka
Welcome to the channel!
Yea help full
No work is done when a person is standing by carrying a load of 50kg for one hour.why
Work is done when it results in a change of energy (kinetic or potential). In this case energy doesn't change.
@@MichelvanBiezen Exactly.Because he is standing.
@@MichelvanBiezen There is no displacement.work=force×displacement
Here's what I remembered before watching the vid. If the component of the direction of motion and the component of the direction of force is perpendicular to each other, no work is done. In order to do work, an object must displace(move) in the same direction as that of the force acting on it. Your arms get tired because your muscles are "doing work against each other." I don't think the problem is whether you're static or have made displacements. It does no work because of the above reasons. Sorry if I'm wrong; that's what I remembered from my textbook.
Cute drawing
😎
Still working on the work, energy and power videos! 🙂
@@MichelvanBiezen You got it, and your replies are always appreciated.
but I did work??? i can't beleive physics??
ok, I don't want to be a dickhead, but wanted to point this out. Obviously the box isn't doing any work, but you should probably look at the work the HANDS and therefore arms and legs are doing to know the work the person is doing, right? I mean obviously there's no work on the box.
It depends what you define as work and what the object is receiving the work. There is no work done on the box (no energy is gained by the box). But even standing there holding the object your muscles will get tired. But holding something up against gravity is strictly speaking not doing work (as defined in physics)