Thank you Sara for letting me know. I'm pleased you found it helpful. I am currently preparing a video on the third law which will be released in the near future. I am interested to know how your teacher used it as part of the lesson? Was it shown to everyone at the same time or part of individual study activity? Regards Mac
Hi Trojan. Good question. One way would be to apply the 50N force to the large block and measure the acceleration achieved. If friction was present the acceleration would be less than the 1m/s2 that was calculated. If you measured the actual acceleration achieved you can calculate the force that was accelerating the block from F=ma. The difference between this calculated force and the 50N applied force is the friction force. Mac
Shubhom, Sorry I can't reply directly to your question as I don't think you have enabled the option on your account to receive comments. The object will only accelerate while a net unbalanced force is applied. So if you push the block it will accelerate while the push is active on the object. As soon as you stop pushing it the block will no longer accelerate. It will continue to have velocity but not acceleration. Mac
Hi Oshyrath. When using equations it is important to define what the variables are. Your point is that in Newton's Second Law of Motion the force we refer to should be the net force. This is calculated by summing the forces on the body under consideration remembering that the direction of the force must be taken into account during this summation. As long as we define F in F=ma as the net force then it is OK to use F=ma. You can also express the equation as you have suggested. Regards Mac
Hi Jason, thanks for your comment. You are right. If the acceleration is zero then it tells us that the forces are in balance. The reason for the exercise was to demonstrate how to use F=ma to calculate acceleration. In the case of the block on the table it is zero. We would use the same analysis approach to calculate a non zero acceleration for the situation where an unbalanced force was active. The zero acceleration case is a good check since it is an obvious result. Regards Mac
The symbol is used to indicate the summation of the forces. It is the upper case letter Sigma from the Greek alphabet. We use this symbol as a quick way to indicate summation when writing equations. Mac
Hi Amie. It is good that you are thinking carefully about the direction of the force since this is important when analysing forces. If you pull an object towards you it will move towards you since it moves in the direction of the pull force. Similarly if you push the object away, it will move away from you as it moves in the direction of the push force. Have I understood your question correctly? Regards Mac.
Hi Hugo, thanks for your comment. You have identified a very important point about using Newton's second law. The force in F=ma is the sum of the forces on the object in a given direction. So vertically you have 49N down (for a mass of 5kg) and 49N up from the table on the block (due to the table supporting the block). The sum of these forces (including their direction) is 49-49=0. See my video on Newton's third law to understand why the table exerts the 49N on the block. Regards Mac
Hi Vonti, I understand why it is confusing. The reason the time is squared is: Velocity is defined as a distanced travelled per unit time. Hence the use of m/s. Acceleration is a change in velocity per unit time. Hence we have change in velocity (m/s) per unit time (s). So acceleration is m/s/s which is m/s^2. For constant acceleration the distance travelled changes with the square of the time. You calculate this using the equation distance= 0.5at^2 assuming the initial velocity is zero. Mac
Hi Guerras thank you for your comment. I understand there is a debate about forces. The aim of these videos is to help students learn about currently accepted concepts. Once students are confident with the concepts they are then able to engage with the more subtle discussion that you have mentioned. I personally find the use of the equation F=ma an effective way to solve engineering problems due to its simple form and accuracy in describing physically observed behaviour. Regards Mac
I had a hard time understanding Newton's 2nd Law of Motion during lecture. The way you were able to simplify everything really helped. Thanks for uploading the awesome video Mac!
Hi Maria, Thanks for your feedback. I'm pleased you found it helpful in your studies. I'm also pleased that you can understand my Scottish accent. I had a good friend who lived in Poland and he sometimes found my accent difficult to understand when talking by phone. Regards Mac.
Thank you so much for uploading this video! It helped me visualize what was happening which helped me understand the concept better. I look forward to your future uploads. :D
Hi Nicolaus. You have good instinct here. However the quantity you are discussing is energy - kinetic energy to be precise. In both cases you have the same car moving with the same velocity and this means it has the same kinetic energy which is defined as 0.5 times mass times velocity squared (similar to what you are suggesting). Acceleration relates to how quickly it achieved this velocity. The larger the acceleration the larger the force required for a given mass of car hence F=ma. Mac
Hi Evan, it has a similar form to F=ma but uses different terms. As an exercise try searching for rotational motion and see if you can find the equation. Mac
Hi Bareerah, I am preparing the video of the third law now. However, since it takes a significant time to create these videos it won't be ready for a few months. You can see screen shots of the video as it progresses at my website. Regards Mac
Well this is better than e-learning websites . Please upload more of these videos I have just one complain to make - the messages which tell us to like and subscribe appear in between the video which disturbs my concentration .
Thank you for this video very much appreciated, you see I come from Poland and English is my second language and although I lived in Great Britain for over 3 years I am still sometimes struggling with understanding the teachers. This was a lot easier for me to understand, very helpful :) Thanks again Maria
Stafus, I assume you mean the upwards acceleration. The inertial mass doesn't change on the moon. What has changed however is the net upwards force on the astronaut and the suit. You need to sum the gravity force and the leg force. So a=F/m will give a higher acceleration for the same mass due to the higher net force. You can't ignore gravity here as it is a force acting on the system and will affect the net force and hence acceleration.
hi mac.....igot a got a doubt..if.. F=m(v-u)÷t...here 't' is showing the time where the force is applied on the object continuesly or it is the time that helps as a reference to determine the 'v'????... in other words...does force needs to be applied continuesly throughout the time or just one push at the start is sufficient to produce accelaration
I also find it much easier to understand when I can visualise the concept. That's one of the reasons I produce my animations. Thanks for your feedback iLearning. Mac
Hi Stafus. You make a good observation. The reason they can move like that on the moon is that the gravitational force on the moon is only approximately 17% of that on earth. Everything weighs less. Not only the suit but also the astronaut. His leg can still generate the same force so the net force upwards (remember to sum the forces) as he jumps is much greater than on earth. The astronaut can therefore jump higher and has a lower acceleration down to the surface than on earth. Mac
Hi Amber, Thank you for your feedback and I'm pleased you liked it. I have just released my new video on Newton's third law and I am in the process of planning my next video. I will take your request into consideration. Mac
Thanks EaszAz. Often the hardest thing is to be first to understand something - even harder to explain it in simple language. Newton did a good job in this regard. The laws are easy to state but have a great deal of significance. We can use them to explain what is happening all around us and use our understanding to design and develop new products to enhance our lives. Mac
@loucris24 Hi and thanks for your feedback. It's great that you are finding my videos helpful in teaching and I hope your students find them an easy way to understand the concepts presented. I have had many requests for a video on the third law and that is planned to be done next once I finish my current video which explains Hooke's law, elasticity and the spring balance. Mac
Hi Mac, Why does the block seem to move in the same direction as the force is applied? Shouldn't the force have pushed the block in the other direction?
Thanks you for your answer. So when physicists are talking about force they are talking about the RESULTS of kinetic energy being applied to an object? Would a more descriptive and long winded explanation in layman’s terms of newton’s second law be: “the result of an object (aka mass) colliding with another object causes the other object to speed up and/or change directions.” So force is the measure of power after the collision and kinetic energy is the measure of power before the collision?
is the spring connected to the box by a magnet, or just connected to the tip by some locking mechanism. and the spring mechanism has to move faster and faster to the right in order to deliver a constant force of 2.5 newtons . i'm a bit confused how the spring mechanism delivers a constant force. you have to hold down the box until the moving arm moves to the right and shows a force of 2.5 pounds, otherwise you enter into a zeno type problem. you move the arm to the right, the box moves to the right instantaneously as well, back and forth, so you can never get it to 2.5 newtons. Thats actually a good question. how fast does the spring arm have to move to the right to maintain a force of 2.5 newtons (assuming the weight of the box is 5 kg). So if you hold the box down, connect the spring to the box and move the spring to the right until it shows 2.5 Newtons, then release, the force on the box is F = m*a , or 2.5 N= 5 kg * a and the acceleration of the box will be 2.5 / 5 meters per second per second, or 0.5 m/s^2. Now the arm of the spring has to accelerate to the right with the same acceleration as the box, they have to move in tandem so that the force of the spring is maintained (since the force of the spring is maintained by that initial gap between box and spring arm). that means after 1 second they both move 0.5 m/s, after 2 second 1 m/s , after 10 seconds 5 m/s . etc. Also to hold the box down while you move the spring 2.5 N to the right, you can place a pin or some kind of stopper to prevent the box from moving. What is interesting is that you have to move the spring faster and faster in order to maintain the constant force. this is also a good example of how to apply a constant force (it is not easy since the object is accelerating and you have to move faster and faster to keep applying the same force).
Hi Xoppa, well done, you have thought carefully about the problem and have a good understanding of the second law. Yes the block is connected to the end of the spring balance by a magnet at the end of the arm and a magnet in the block. The easy way to think about the motion of the arm to provide a force of 2.5N is to first think about acceleration of the block. If the block has a force of 2.5N applied, the acceleration of the block (with a mass of 5kg) is, as you have calculated, 0.5m/s^2. So if you now accelerate the arm at 0.5m/s^2 and connect the block by a spring, the two will, after a short period of time, accelerate at 0.5m/s^2. Why? Well In order for the spring to generate a force it has to extend. The more it extends the larger the force. So when the arm starts to accelerate, the spring applies a small but increasing force as the spring extends more and more. This means that initially the block has a small but increasing acceleration. During this time the distance between the arm and the block is increasing, and the spring is extending and therefore the force on the block is increasing. Eventually there will be situation where the force on the block is the force required to accelerate it at the same acceleration as the arm. Since we have set this to be 0.5m/s^2, then the force on the block must be 2.5N from Newton's second law (assuming a mass of the block of 5kg). The reason why this balance must occur is that if the block were to be accelerating faster than the arm, then the distance between the arm and the block would decrease and the force from the spring would decrease, which in turn would result in the block slowing down so that the distance between the arm and the block would increase again with an associated increase in the force. Eventually there will be the balance point where the force is 2.5N which will accelerate the block at 0.5m/s^2, exactly the same as the acceleration of the arm. All we have to do is control the acceleration of the arm to control the force on the block. Regards Mac.
Will the force be zero if the block is moving with constant velocity (ie., zero acceleraton). I mean the block is moving equal distance for every interval of time. let us say 5kg block moves 1 meter distance for every second. In this case, what will happen to the force?
Great question Aravindh. You are correct. If the block is moving with constant velocity in a straight line it has zero acceleration. Using F=ma, with acceleration equal to zero, gives the net force on the block equal to zero. Note that I said the net force. There may be many forces on the block, but if it has zero acceleration in a given direction then all the forces in that direction must balance. This gives zero net force which gives zero acceleration in that direction. Regards Mac.
learnwithmac My doubt is, If the force is zero for zero acceleration of the body of mass 5kg moving horizontally on a frictionless floor with constant velocity (1m/s), which force made the body to move? what is the magnitude of the force which made the 5kg object to move with constant velocity of 1m/s?
Aravindh, you can't work that out based on observing the object once it is moving with constant velocity. You can only work it out by observing the acceleration of the object. If you measure the object's acceleration you can work out the force using the equation F=ma. The point which you may be finding difficult is that there is no single answer. It depends on the size of the acceleration. If the acceleration was very small then the applied force was small. If the acceleration was large it was a large force. No matter what the value of the acceleration it will eventually change the velocity of the object such that it can reach 1m/s. It takes longer with a smaller acceleration (therefore a smaller force). Mac
Hi Mac, very explaining video. I'm just wondering. Since the cube was moving 1/ms (meters per second) how come the distance the cube travels increases for every second that passes? It looks like the Meter per second increases for each second. Please explain.
Hi Daniel, you are correct. The distance is increasing because the block is accelerating at (1 meter per second) per second. This means the velocity is increasing every second by 1 meter per second. You thought the block was moving with a constant velocity of 1 meter per second but it was actually accelerating so the velocity is not constant. This means that it will travel further each second due to the increasing velocity. Mac
Hi Bobby, I can't reply to you directly as you don't have that option selected on your account. Thanks for letting me know about your test result. Well done. Mac
@kalumizkool Hi, I'm currently in the process of producing a video explaining the spring balance in more detail. After that I will start the video on the third law. Mac
Thank you so much mac! You really are good at explaining, cause now I understand it, good thing I stumbled here, cause this year there is a pandemic so we could'nt go to class and because of that no one can teach us and so they send us modules. And I really can't understand a topic unless it is being teached so thank you so much!!
Please help me professor Mac, please answer my question: A 100N weight rest on a 30 degrees inclined plane.Neglecting friction,how much pull must one exert to bring the weight up the plane?
Thank you Sara for letting me know. I'm pleased you found it helpful. I am currently preparing a video on the third law which will be released in the near future. I am interested to know how your teacher used it as part of the lesson? Was it shown to everyone at the same time or part of individual study activity?
Regards
Mac
Hi Trojan. Good question. One way would be to apply the 50N force to the large block and measure the acceleration achieved. If friction was present the acceleration would be less than the 1m/s2 that was calculated. If you measured the actual acceleration achieved you can calculate the force that was accelerating the block from F=ma. The difference between this calculated force and the 50N applied force is the friction force. Mac
Shubhom, Sorry I can't reply directly to your question as I don't think you have enabled the option on your account to receive comments. The object will only accelerate while a net unbalanced force is applied. So if you push the block it will accelerate while the push is active on the object. As soon as you stop pushing it the block will no longer accelerate. It will continue to have velocity but not acceleration. Mac
>> net unbalanced force i
Thank you so much Mac..I understood very well the concept of second law....
I appreciate the huge amount of work that has gone into this video- congratulations.
there should be more of these videos! i love them!
Hi Oshyrath. When using equations it is important to define what the variables are. Your point is that in Newton's Second Law of Motion the force we refer to should be the net force. This is calculated by summing the forces on the body under consideration remembering that the direction of the force must be taken into account during this summation. As long as we define F in F=ma as the net force then it is OK to use F=ma. You can also express the equation as you have suggested. Regards Mac
Hi Jason, thanks for your comment. You are right. If the acceleration is zero then it tells us that the forces are in balance. The reason for the exercise was to demonstrate how to use F=ma to calculate acceleration. In the case of the block on the table it is zero. We would use the same analysis approach to calculate a non zero acceleration for the situation where an unbalanced force was active. The zero acceleration case is a good check since it is an obvious result. Regards Mac
Thanks for your feedback Manik. I will look at the message and see if I can move it to a less distracting time during the video. Mac
Thank you sir for the reply and the clarity , conveinced almost
Thank you for this video! It helped me understand Newtons second law of motion much easier.
Thank you for your feedback Eman. I'm pleased you like them.
Regards Mac
The symbol is used to indicate the summation of the forces. It is the upper case letter Sigma from the Greek alphabet. We use this symbol as a quick way to indicate summation when writing equations.
Mac
This is very informative! I plan to show this to my high school students to increase understanding!
Thank you for the explanation, professor. (: Loved the video and hope you make more!
Hi Amie. It is good that you are thinking carefully about the direction of the force since this is important when analysing forces. If you pull an object towards you it will move towards you since it moves in the direction of the pull force. Similarly if you push the object away, it will move away from you as it moves in the direction of the push force. Have I understood your question correctly?
Regards Mac.
Thanks alot mac, really helpful while im studying for exams. Really appreciate it.
Thanks Dias. I will take your suggestions into consideration as I plan the next video.
Regards, Mac
Hi Hugo, thanks for your comment. You have identified a very important point about using Newton's second law. The force in F=ma is the sum of the forces on the object in a given direction. So vertically you have 49N down (for a mass of 5kg) and 49N up from the table on the block (due to the table supporting the block). The sum of these forces (including their direction) is 49-49=0. See my video on Newton's third law to understand why the table exerts the 49N on the block. Regards Mac
This is amazing, it's much clearer than my text books :D Thank you
Excellent presentation and easy to follow. I am looking forward to your other videos. Thanks for making it.
Great video thankyou .
I really understood the Newton second law.
@singaporeoracherd Thanks for your feedback. I'm pleased you found it helpful and I wish you every success in your studies.
Regards Mac
Thanks for your help.I will come with more questions on physics.
Hi Vonti, I understand why it is confusing. The reason the time is squared is: Velocity is defined as a distanced travelled per unit time. Hence the use of m/s. Acceleration is a change in velocity per unit time. Hence we have change in velocity (m/s) per unit time (s). So acceleration is m/s/s which is m/s^2. For constant acceleration the distance travelled changes with the square of the time. You calculate this using the equation distance= 0.5at^2 assuming the initial velocity is zero. Mac
Hi Guerras thank you for your comment. I understand there is a debate about forces. The aim of these videos is to help students learn about currently accepted concepts. Once students are confident with the concepts they are then able to engage with the more subtle discussion that you have mentioned. I personally find the use of the equation F=ma an effective way to solve engineering problems due to its simple form and accuracy in describing physically observed behaviour. Regards Mac
This is so helpful. Thanks a lot, Mac!
Koen, I use the open source program called Blender. The link to the site for download is shown in the credits at the end of my video.
Mac
Whos that
I had a hard time understanding Newton's 2nd Law of Motion during lecture. The way you were able to simplify everything really helped. Thanks for uploading the awesome video Mac!
seanbarrett503 Thanks for your feedback Sean.
Well done
Thanks!! this makes a great help on understanding law of acceleration!
Thank you Mac this was very helpful. I got a 100 on my test thanks to you! ;) just wanted to say thanks.
How life my guy
Congos
:) I slept through class and thanks to this video I understand the lesson I missed today :)
Great video.
Important from the outset to state that the f in f=ma is the Resultant or unbalanced Force on an object.
Hi Maria, Thanks for your feedback. I'm pleased you found it helpful in your studies. I'm also pleased that you can understand my Scottish accent. I had a good friend who lived in Poland and he sometimes found my accent difficult to understand when talking by phone. Regards Mac.
that accent took me off guard so much
Indian accent
Scottesh
JazziHD Ireland’s
Thank you so much for uploading this video! It helped me visualize what was happening which helped me understand the concept better. I look forward to your future uploads. :D
Tanks for your explanation, it is really the best on the internet.
Thanks a lot! This really helped me with a project I'm doing!
Hi Nicolaus. You have good instinct here. However the quantity you are discussing is energy - kinetic energy to be precise. In both cases you have the same car moving with the same velocity and this means it has the same kinetic energy which is defined as 0.5 times mass times velocity squared (similar to what you are suggesting). Acceleration relates to how quickly it achieved this velocity. The larger the acceleration the larger the force required for a given mass of car hence F=ma.
Mac
Hi Pescarlerik. Thanks for your feedback. Yes I am originally from Glasgow area and studied at Glasgow University.
Mac
I got a 100 on my physics test thanks to this I love u
Well done Naram.
Mac
Hi Evan, it has a similar form to F=ma but uses different terms. As an exercise try searching for rotational motion and see if you can find the equation.
Mac
Hi Bareerah, I am preparing the video of the third law now. However, since it takes a significant time to create these videos it won't be ready for a few months. You can see screen shots of the video as it progresses at my website.
Regards Mac
Hi Evan, I don't understand your question. What do you refer to when you say revolutions? Revolutions of what?
Mac
good stuff, do more,I got an A star for my exam on forces:)
+Abisai Yogarajan Thanks for you feedback. Good job on your exam results. Mac
Thank you Mac! This video is very helpful to students like me.
Thanks for your feedback Albert.
Well this is better than e-learning websites . Please upload more of these videos
I have just one complain to make - the messages which tell us to like and subscribe appear in between the video which disturbs my concentration .
Absolutely amazing! Thank You very much for this information! This was explained much better than my prof explained it.
🤣
Thank you for this video very much appreciated, you see I come from Poland and English is my second language and although I lived in Great Britain for over 3 years I am still sometimes struggling with understanding the teachers. This was a lot easier for me to understand, very helpful :)
Thanks again Maria
Its so amazing, I'm gonna use your video learning as my teaching source
Thanks for your feedback Muhammad.
Mac
Stafus, I assume you mean the upwards acceleration. The inertial mass doesn't change on the moon. What has changed however is the net upwards force on the astronaut and the suit. You need to sum the gravity force and the leg force. So a=F/m will give a higher acceleration for the same mass due to the higher net force. You can't ignore gravity here as it is a force acting on the system and will affect the net force and hence acceleration.
hi mac.....igot a got a doubt..if.. F=m(v-u)÷t...here 't' is showing the time where the force is applied on the object continuesly or it is the time that helps as a reference to determine the 'v'????... in other words...does force needs to be applied continuesly throughout the time or just one push at the start is sufficient to produce accelaration
Just what I thought for decades . Acceleration needs a constant increase in force in order to cover larger distances in the same time . Right ?
Hey Mac, will you please talk me through calculating tension on a smooth horizontal plane?
It seems he never replied
@@philmccrack9862 😂😂😂 I can't even remember watching this. Got a maths degree out of it though
I Likd it. My doubts are all cleared, But what is Linear Momentum??
Please answer...
Nice explanation
Thank you Mac. Your videos are the best!
I also find it much easier to understand when I can visualise the concept. That's one of the reasons I produce my animations. Thanks for your feedback iLearning.
Mac
very good video presentation it entertain yourself and you can learn more how did you do that thing Professor Mac?
Hi Rogelio, thanks for your comment. Have a look at my blog to see how I create the animations. Here is the link: www.learnwithmac.com
Hi Stafus. You make a good observation. The reason they can move like that on the moon is that the gravitational force on the moon is only approximately 17% of that on earth. Everything weighs less. Not only the suit but also the astronaut. His leg can still generate the same force so the net force upwards (remember to sum the forces) as he jumps is much greater than on earth. The astronaut can therefore jump higher and has a lower acceleration down to the surface than on earth.
Mac
Hi Amber, Thank you for your feedback and I'm pleased you liked it. I have just released my new video on Newton's third law and I am in the process of planning my next video. I will take your request into consideration.
Mac
Hi Bareerah, I will be preparing the video on the third law in the near future.
Regards Mac
BEST VIDEO EVER WATCHED!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!CONCEPT CLEARING VIDEO
+Sanjay Verma Thanks for your feedback Sanjay
Nice explanations. Now I'm clear about force, mass and acceleration. Thank you Mac.
Thanks for your feedback Abdur. Regards Mac
useful for students
Thanks a lot prof Mac
Thanks for the feedback Beshoo. I'm glad you found it helpful. Mac
Thanks EaszAz. Often the hardest thing is to be first to understand something - even harder to explain it in simple language. Newton did a good job in this regard. The laws are easy to state but have a great deal of significance. We can use them to explain what is happening all around us and use our understanding to design and develop new products to enhance our lives. Mac
its helpfull thanks
mac
Hi
I am trying to learn for a science test and this helping me study thank you so so much!!
youaretheonlyone Thanks for your feedback. Best of luck in your test. Regards Mac
I'm pleased you found it helpful Alma.
Mac
@loucris24 Hi and thanks for your feedback. It's great that you are finding my videos helpful in teaching and I hope your students find them an easy way to understand the concepts presented. I have had many requests for a video on the third law and that is planned to be done next once I finish my current video which explains Hooke's law, elasticity and the spring balance. Mac
Hi Mac,
Why does the block seem to move in the same direction as the force is applied? Shouldn't the force have pushed the block in the other direction?
Thanks you for your answer. So when physicists are talking about force they are talking about the RESULTS of kinetic energy being applied to an object? Would a more descriptive and long winded explanation in layman’s terms of newton’s second law be: “the result of an object (aka mass) colliding with another object causes the other object to speed up and/or change directions.” So force is the measure of power after the collision and kinetic energy is the measure of power before the collision?
is the spring connected to the box by a magnet, or just connected to the tip by some locking mechanism. and the spring mechanism has to move faster and faster to the right in order to deliver a constant force of 2.5 newtons
. i'm a bit confused how the spring mechanism delivers a constant force. you have to hold down the box until the moving arm moves to the right and shows a force of 2.5 pounds, otherwise you enter into a zeno type problem. you move the arm to the right, the box moves to the right instantaneously as well, back and forth, so you can never get it to 2.5 newtons.
Thats actually a good question. how fast does the spring arm have to move to the right to maintain a force of 2.5 newtons (assuming the weight of the box is 5 kg).
So if you hold the box down, connect the spring to the box and move the spring to the right until it shows 2.5 Newtons, then release, the force on the box is F = m*a , or 2.5 N= 5 kg * a
and the acceleration of the box will be 2.5 / 5 meters per second per second, or 0.5 m/s^2. Now the arm of the spring has to accelerate to the right with the same acceleration as the box, they have to move in tandem so that the force of the spring is maintained (since the force of the spring is maintained by that initial gap between box and spring arm). that means after 1 second they both move 0.5 m/s, after 2 second 1 m/s , after 10 seconds 5 m/s . etc.
Also to hold the box down while you move the spring 2.5 N to the right, you can place a pin or some kind of stopper to prevent the box from moving. What is interesting is that you have to move the spring faster and faster in order to maintain the constant force. this is also a good example of how to apply a constant force (it is not easy since the object is accelerating and you have to move faster and faster to keep applying the same force).
Hi Xoppa, well done, you have thought carefully about the problem and have a good understanding of the second law. Yes the block is connected to the end of the spring balance by a magnet at the end of the arm and a magnet in the block.
The easy way to think about the motion of the arm to provide a force of 2.5N is to first think about acceleration of the block. If the block has a force of 2.5N applied, the acceleration of the block (with a mass of 5kg) is, as you have calculated, 0.5m/s^2. So if you now accelerate the arm at 0.5m/s^2 and connect the block by a spring, the two will, after a short period of time, accelerate at 0.5m/s^2. Why? Well In order for the spring to generate a force it has to extend. The more it extends the larger the force. So when the arm starts to accelerate, the spring applies a small but increasing force as the spring extends more and more. This means that initially the block has a small but increasing acceleration. During this time the distance between the arm and the block is increasing, and the spring is extending and therefore the force on the block is increasing. Eventually there will be situation where the force on the block is the force required to accelerate it at the same acceleration as the arm. Since we have set this to be 0.5m/s^2, then the force on the block must be 2.5N from Newton's second law (assuming a mass of the block of 5kg). The reason why this balance must occur is that if the block were to be accelerating faster than the arm, then the distance between the arm and the block would decrease and the force from the spring would decrease, which in turn would result in the block slowing down so that the distance between the arm and the block would increase again with an associated increase in the force. Eventually there will be the balance point where the force is 2.5N which will accelerate the block at 0.5m/s^2, exactly the same as the acceleration of the arm. All we have to do is control the acceleration of the arm to control the force on the block. Regards Mac.
Thank you for this very clear and informative demonstration.
Thanks for your feedback Jan.
Mac
Awesome professor Mac beautifully explained, easily understood!!!!!!!!
So clear and concise, thanks Mac!
David Gordon Thanks for the feedback David.Mac
Well done !!! best explanation i have seen
Will the force be zero if the block is moving with constant velocity (ie., zero acceleraton). I mean the block is moving equal distance for every interval of time. let us say 5kg block moves 1 meter distance for every second. In this case, what will happen to the force?
Great question Aravindh. You are correct. If the block is moving with constant velocity in a straight line it has zero acceleration. Using F=ma, with acceleration equal to zero, gives the net force on the block equal to zero. Note that I said the net force. There may be many forces on the block, but if it has zero acceleration in a given direction then all the forces in that direction must balance. This gives zero net force which gives zero acceleration in that direction. Regards Mac.
learnwithmac
My doubt is, If the force is zero for zero acceleration of the body of mass 5kg moving horizontally on a frictionless floor with constant velocity (1m/s), which force made the body to move?
what is the magnitude of the force which made the 5kg object to move with constant velocity of 1m/s?
Aravindh, you can't work that out based on observing the object once it is moving with constant velocity. You can only work it out by observing the acceleration of the object. If you measure the object's acceleration you can work out the force using the equation F=ma. The point which you may be finding difficult is that there is no single answer. It depends on the size of the acceleration. If the acceleration was very small then the applied force was small. If the acceleration was large it was a large force. No matter what the value of the acceleration it will eventually change the velocity of the object such that it can reach 1m/s. It takes longer with a smaller acceleration (therefore a smaller force). Mac
Hi Mac, very explaining video. I'm just wondering. Since the cube was moving 1/ms (meters per second) how come the distance the cube travels increases for every second that passes? It looks like the Meter per second increases for each second. Please explain.
Hi Daniel, you are correct. The distance is increasing because the block is accelerating at (1 meter per second) per second. This means the velocity is increasing every second by 1 meter per second. You thought the block was moving with a constant velocity of 1 meter per second but it was actually accelerating so the velocity is not constant. This means that it will travel further each second due to the increasing velocity. Mac
Thank you!!
plzzz make videos on strength of material and thermodynamics and fluid mechanics and hydrolycis
Excellent teacher! Unfortunately my physics teacher doesn't really explain things well. But I learn so quickly from watching your videos!
These videos are really good! I like how you also explain the maths side of things. Keep up the good work!!
Thanks for your feedback K.
Mac
Hi Bobby, I can't reply to you directly as you don't have that option selected on your account. Thanks for letting me know about your test result. Well done. Mac
Hi mike I like how explain 2nd law
Learnwithmac is the best
@kalumizkool Hi, I'm currently in the process of producing a video explaining the spring balance in more detail. After that I will start the video on the third law. Mac
Thanks for your time that you have explained us
Everyone is probably here for a homework but I'm here because my teacher isn't good at explaining and I have a quiz in about 15 hours
Nice work,Professor....
This video help me a lots to understand more about the 2nd law of motion thank you so much
Thanks for your comment Eurika. I'm pleased you found it helpful.
Mac
When you speak of "applied force" , should it be written as applied directly to the mass ; such as FxM ???
EGMAG F(Applied)=mg
The applied force results in an acceleration. It is applied directly to the object.
It is applied directly to the object.
applied directly to the object.
Also your video was excellent hope you make more
Thank you! Well presented.
thanks. very useful information and nice presenting.
very clear illustration
Thanks for your feedback Senen. I'm pleased you found it easy to understand. Mac
Thanks alot Mac.... You really helped alot explaining the second law..!!
+Stephan van Zyl Thanks for your feedback Stephan. I'm pleased you found it helpful. Mac
Thank you so much mac! You really are good at explaining, cause now I understand it, good thing I stumbled here, cause this year there is a pandemic so we could'nt go to class and because of that no one can teach us and so they send us modules. And I really can't understand a topic unless it is being teached so thank you so much!!
Thanks for your feedback Kirstein
Please help me professor Mac, please answer my question: A 100N weight rest on a 30 degrees inclined plane.Neglecting friction,how much pull must one exert to bring the weight up the plane?
Very informative video. Thanks for sharing.
+Diane Blackford Thanks for your feedback Diane. Mac