*Please click video title headings to play video if video does not load properly.
Newton's Three Laws of Motion
1st Law: An object will remain at rest or in a uniform motion in a straight line unless acted upon by an unbalanced force.
From 1st Law, we have the idea of inertia which is the resistance of an object to a change in its state of motion or rest.
<We can understand inertia better by imagining that it is difficult to push and start a car which is at rest to move. Similarly, a car which is moving at constant velocity will be very difficult to stop it moving. Of course, we invented car engines and brakes to make the job of starting and stopping a car easy. However, imagine the car engine breaks down or the brakes refuse to work, then we will really appreciate inertia again since we cannot easily start our car moving or stop it while it is still moving in such cases.>
2nd Law: The resultant force on an object is proportional to the product of its mass and acceleration.
From 2nd Law, we have the formula F = ma , F is resultant force, m is mass, a is acceleration.
3rd Law: For every action, there is an equal and opposite reaction.
From 1st Law, we have the idea of inertia which is the resistance of an object to a change in its state of motion or rest.
<We can understand inertia better by imagining that it is difficult to push and start a car which is at rest to move. Similarly, a car which is moving at constant velocity will be very difficult to stop it moving. Of course, we invented car engines and brakes to make the job of starting and stopping a car easy. However, imagine the car engine breaks down or the brakes refuse to work, then we will really appreciate inertia again since we cannot easily start our car moving or stop it while it is still moving in such cases.>
2nd Law: The resultant force on an object is proportional to the product of its mass and acceleration.
From 2nd Law, we have the formula F = ma , F is resultant force, m is mass, a is acceleration.
3rd Law: For every action, there is an equal and opposite reaction.
Worked problems on Newton's Second Law (F = ma) (Part 1)
In this video, the first problem is a simple and common one. The second problem is more complicated which may not be asked during O levels. This complicated type of problem is more relevant to students from integrated program schools. However, it is also good that O level students go through this complicated problem to train your mind to think more indepth as there is application of forming of equations which is relevant to mathematics to solve this problem. Afterall, all these problems are just depending on the physics formula F = ma to solve them.
Worked problems on Newton's Second Law (F = ma) (Part 2)
This video is a continuation to the part 1 video above. The complicated problem from the earlier part 1 video can be solved by another simpler way which does not involve forming of equations (see this video).
The next problem in this video is a terminal velocity problem. Take note that when an object is falling at constant terminal velocity, its acceleration MUST BE 0 ms-² since it is no longer accelerating at constant terminal velocity. At this instant, the object's force of weight acting downwards MUST BE EQUAL to the force of air resistance acting upwards. The object's resultant force will therefore be 0 N as well. The working is as follows.
Resultant force of person falling = Force of weight - Force of air resistance ,
<the downwards direction of weight is chosen to be +ve while upwards direction of air resistance is chosen to be -ve>
Since Force of weight and Force of air resistance ARE EQUAL at terminal velocity, these two forces subtract to give resultant force to be 0 N.
As F = ma , assuming the person's mass is 60 Kg, when resultant force F = 0 N at terminal velocity,
0 = 60 X a
a = 0 ms-²
Thus, a which is acceleration must be 0 ms-² at terminal velocity. Recall that terminal velocity is a constant velocity reached while falling. So, there must not be any acceleration if the person is now moving at constant terminal velocity.
We arrive at the conclusion for this kind of falling problem involving terminal velocity, there are two important concepts which can be simply understood.
First concept: Resultant force on object falling at terminal velocity is 0 N.
Second concept: Acceleration of object falling at terminal velocity is also 0 ms-² since it is no longer accelerating.
The next problem in this video is a terminal velocity problem. Take note that when an object is falling at constant terminal velocity, its acceleration MUST BE 0 ms-² since it is no longer accelerating at constant terminal velocity. At this instant, the object's force of weight acting downwards MUST BE EQUAL to the force of air resistance acting upwards. The object's resultant force will therefore be 0 N as well. The working is as follows.
Resultant force of person falling = Force of weight - Force of air resistance ,
<the downwards direction of weight is chosen to be +ve while upwards direction of air resistance is chosen to be -ve>
Since Force of weight and Force of air resistance ARE EQUAL at terminal velocity, these two forces subtract to give resultant force to be 0 N.
As F = ma , assuming the person's mass is 60 Kg, when resultant force F = 0 N at terminal velocity,
0 = 60 X a
a = 0 ms-²
Thus, a which is acceleration must be 0 ms-² at terminal velocity. Recall that terminal velocity is a constant velocity reached while falling. So, there must not be any acceleration if the person is now moving at constant terminal velocity.
We arrive at the conclusion for this kind of falling problem involving terminal velocity, there are two important concepts which can be simply understood.
First concept: Resultant force on object falling at terminal velocity is 0 N.
Second concept: Acceleration of object falling at terminal velocity is also 0 ms-² since it is no longer accelerating.
Simple typical worked problem on Newton's Second Law (F= ma)
In this video, it provides a simple and typical problem which can be solved by finding resultant force first, and then followed by acceleration. The direction to the right in this problem is chosen to be +ve while to the left is -ve.
Thus, Resultant force = 250 N - 50 N = 200 N.
Then, using F = ma,
200 = (50) a
a = 4 ms-²
Acceleration of the hockey gear is 4 ms-² to the right since to the right is +ve direction.
Please follow on to some extension questions below not mentioned in the video.
Qn: Is the hockey gear moving with constant velocity?
Ans: Obviously not since there is an acceleration of 4 ms-².
Qn: If the hockey gear later moves at a constant velocity, what will be its acceleration?
Ans: Its acceleration will become 0 ms-².
Qn: What will now be the required forward pulling force to the right to allow this constant velocity to be reached (assuming frictional force to the left is still kept at 50 N)?
Ans: The pulling force will have to be also 50 N but to the right so that the resultant force on the hockey gear is 0 N (Resultant force = 50 - 50 = 0 N) meaning the acceleration is also 0 ms-² since F= ma. Thus, with no acceleration, the hockey gear can now move at constant velocity to the right.
**Note that some students have the misconception that when resultant force is 0 N, the object must always be at rest. In this extension question, it shows us that resultant force is 0 N can also mean that the object may be moving at constant velocity with no acceleration! So, resultant force is 0 N can actually have two possible scenarios (either the object is at rest or moving at constant velocity e.g. in this extension question)!
Thus, Resultant force = 250 N - 50 N = 200 N.
Then, using F = ma,
200 = (50) a
a = 4 ms-²
Acceleration of the hockey gear is 4 ms-² to the right since to the right is +ve direction.
Please follow on to some extension questions below not mentioned in the video.
Qn: Is the hockey gear moving with constant velocity?
Ans: Obviously not since there is an acceleration of 4 ms-².
Qn: If the hockey gear later moves at a constant velocity, what will be its acceleration?
Ans: Its acceleration will become 0 ms-².
Qn: What will now be the required forward pulling force to the right to allow this constant velocity to be reached (assuming frictional force to the left is still kept at 50 N)?
Ans: The pulling force will have to be also 50 N but to the right so that the resultant force on the hockey gear is 0 N (Resultant force = 50 - 50 = 0 N) meaning the acceleration is also 0 ms-² since F= ma. Thus, with no acceleration, the hockey gear can now move at constant velocity to the right.
**Note that some students have the misconception that when resultant force is 0 N, the object must always be at rest. In this extension question, it shows us that resultant force is 0 N can also mean that the object may be moving at constant velocity with no acceleration! So, resultant force is 0 N can actually have two possible scenarios (either the object is at rest or moving at constant velocity e.g. in this extension question)!
Explain why a car starting from rest reaches a maximum constant speed when the engine is operating at its maximum power.
As the car starts from rest, the forward driving force is greater than the total resistive forces of friction and air resistance acting against the motion. The car experiences a resultant force in forward direction. Since resultant force = mass X acceleration, the car accelerates and its speed increases.
As the car engine is operating at its maximum power, the car is travelling with a maximum constant forward driving force. The total resistive forces of friction and air resistance increases to oppose motion. The resultant force of car in forward direction decreases. Thus, acceleration of car decreases. The car increases its speed at a decreasing rate.
After sometime, the total resistive forces of friction and air resistance increases until they are equal to the forward maximum constant driving force. The resultant force thus becomes 0 N and there is no net resultant force on the car. Thus, there is no acceleration of the car and it travels with a maximum constant terminal velocity in forward direction.