In the previous section we saw Newton's first law. We also saw the basics of inertia. In this section, we will see Newton's second law.
■ The first law deals with the simplest case:
• The net force on a body is zero
♦ As a result, a body at rest remains at rest
♦ As a result, a body in motion remains in motion
■ The second law deals with the general situation:
• The net force on the body is not zero
♦ That is., a net force is indeed acting on the body.
Ans: The 'state of rest' will change. That is., body will begin to move
• To move means, it has a velocity
♦ This velocity may be uniform or non-uniform
• Consider the instantaneous velocity at any instant after the motion has started. Let it be v
• The initial velocity was zero
• The new velocity of v cannot be attained with out acceleration from zero value
■ So we can write:
Net force acts on a body at rest
⇒ Velocity change from zero to v
⇒ The body experienced acceleration
B. Body in uniform motion
• What happens when a net force act on a body in uniform motion?
Ans: The ‘state of uniform motion’ will change
• This change implies that velocity changes
• Consider the instantaneous velocity at any instant after the change has occurred. Let it be v2
• Let the initial velocity be v1
• The new velocity of v2 cannot be attained with out acceleration/retardation from v1
■ So we can write:
Net force acts on a body in uniform motion
⇒ Velocity change from v1 to v2
⇒ The body experienced acceleration/retardation
• We want the relation between this 'acceleration' and the 'net external force'
• For that, we will do some experiments:
Experiment 1:
1. Consider a car on a level road. Fig.5.4(a) below:
• It is in a state of rest. We want to push it manually to the right
2. For that, we have to apply force. Let us apply force in a systematic way.
• Because, we want to take 'time' also into consideration.
3. Let a force F1 be applied from left to right
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't1' in the stop-watch will be 0
4. The car will not move at the same instant when F1 is applied. It will take some time to start moving
• If the car is big it will take a longer time
• If the car is small it will take a only a shorter time.
• Once it start moving, it's velocity will go on increasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F1 is applied
(ii) The instant when the car attains a velocity of say 2 ms-1
(we can fix any convenient value for the velocity. 2 ms-1 is only an example)
6. To measure Δt, we must carefully note down the stop-watch reading 't2' at the instant when the speedometer reading reaches 2 ms-1
• Let t2 = t
• Then Δt = (t2 - t1) = (t-0) = t s
7. So we can write:
• A force F1 is required to push a car from rest, to move it with a velocity of 2 ms-1
• We will write the steps:
1. Consider a truck on the same road in experiment 1
• It is in a state of rest. We want to push it manually to the right
2. For that, we have to apply force. Let us apply force in a systematic way.
• Because, we want to take 'time' also into consideration.
3. Let a force F be applied from left to right
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't1' in the stop-watch will be 0
4. The truck will not move at the same instant when F is applied. It will take some time to start moving
• Once it starts moving, it's velocity will go on increasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F is applied
(ii) The instant when the truck attains a velocity of say 2 ms-1
6. To measure Δt, we must carefully note down the stop-watch reading 't2' at the instant when the speedometer reading is 2 ms-1
• Let t2 = t
• Then Δt = (t2 - t1) = (t-0) = t s
7. But now there is a problem.
• We want both time and velocity to be the same as in experiment 1.
• That is., we want to attain the velocity 2 ms-1 in the same interval 't' obtained in experiment 1
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the truck to rest and start pushing it again
8. We do the trials until the following two conditions are satisfied
(i) The truck attains a velocity of 2 ms-1.
(ii) This velocity is attained in the same time duration 't' as in experiment 1
• The force F which satisfies the both two conditions can be noted down as F2
■ We can write:
• A force F2 is required to push the truck from rest, to move it with a velocity of 2 ms-1.
• The time required for this velocity change (from zero to 2 ms-1) is the same t s as in experiment 1
■ We will find that F1 < F2.
• The truck obviously has greater mass than the car
■ We can write: When mass increases, greater force is required.
Let us do another set of two experiments to confirm this:
Experiment 3:
1. Consider a 'trolley carrying a mass' on a level road. See fig.5.5(a) below:
• It is in a state of uniform motion towards the right.
• It's velocity is 1.5 ms-1.
• We want to bring it to a stop
2. For that, we have to apply force towards the left. Let us apply force in a systematic way.
• Because, we want to take 'time' also into consideration.
3. Let a force F3 be applied from right to left
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't1' in the stop-watch will be 0
4. The trolley will not stop at the same instant when F3 is applied. It will take some time to stop
• If the trolley contains more mass, it will take a longer time
• If the trolley contains less mass, it will take a only a shorter time.
• Once it start to slow down, it's velocity will go on decreasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F3 is applied
(ii) The instant when the trolley attains a velocity of 0 ms-1
6. To measure Δt, we must carefully note down the stop-watch reading 't2' at the instant when the trolley comes to rest
• Let t2 = t
• Then Δt = (t2 - t1) = (t-0) = t s
7. So we can write:
• A force F3 is required to bring the trolley to rest.
• The time required for this velocity change (from 1.5 ms-1 to zero) is t s
• We will write the steps:
1. Consider the 'trolley with increased mass' on the same floor as in experiment 3
• It is in a state of uniform motion.
• It's velocity is 1.5 ms-1.
• We want to bring it to a stop
2. For that, we have to apply force towards the left. Let us apply force in a systematic way.
• Because, we want to take 'time' also into consideration.
3. Let a force F be applied from right to left
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't1' in the stop-watch will be 0
4. The trolley will not stop at the same instant when F is applied. It will take some time to stop
• Once it start to slow down, it's velocity will go on decreasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F3 is applied
(ii) The instant when the trolley attains a velocity of 0 ms-1
6. To measure Δt, we must carefully note down the stop-watch reading 't2' at the instant when the trolley comes to rest
• Let t2 = t
• Then Δt = (t2 - t1) = (t-0) = t s
7. But now there is a problem.
• We want both time and velocity to be the same as in experiment 1.
• That is., we want to attain the velocity 0 ms-1 in the same interval 't' obtained in experiment 3
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the trolley to 'uniform motion at 1.5 ms-1' and try to stop it again
8. We do the trials until the following two conditions are satisfied
(i) The trolley attains a velocity of 0 ms-1.
(ii) This velocity is attained in the same time duration 't' as in experiment 3
• The force F which satisfies the both two conditions can be noted down as F4
■ We can write:
• A force F4 is required to bring the trolley to rest.
• The time required for this velocity change (from 1.5 ms-1 to zero) is the same t s obtained in experiment 3
■ We will find that F3 < F4.
• The trolley in experiment 4 obviously has greater mass than that in experiment 1
■ We can write: When mass increases, greater force is required.
Let us do one more 'set of two experiments' to confirm this. These are simple experiments:
Experiment 5:
1. Drop a small stone from the top of a building
2. Let a person standing at the foot of the building catch it
3. The following points should be noted while catching:
♦ The person must wear a pair of good quality work gloves. This is to avoid injury.
♦ The person must not lower his hands while making the catch
4. Let the force experienced by the person be F5
Experiment 6:
1. Drop a stone which is a little bigger (than the one used in experiment 5) from the top of the same building
2. Let the person standing at the foot of the building catch it
3. The following points should be noted this time also:
♦ The person must wear a pair of good quality work gloves. This is to avoid injury.
♦ The person must not lower his hands while making the catch
4. Let the force experienced by the person be F6
■ It will be found that F5 < F6
The following points may be noted:
(i) The bigger stone has greater mass
(ii) The height is same in both experiments
♦ So the 'velocity at the time of catch' will be the same in both cases
(iii) The person does not lower his hands
♦ So the 'time of application of the force' is same in both cases
• When mass increases, greater force is required.
• When mass decreases, lesser force is required.
In other words:
■ Force is directly proportional to mass
• symbolically we write this as: F∝ m
♦ Where 'F' represents the force and 'm' represents the mass
♦ The symbol '∝' stands for 'is proportional to'
• We obtained it by changing the mass while keeping velocity and time constant
• In the next section we will obtain the relation between velocity and force
♦ For that, we will change the velocity while keeping mass and time constant
■ The first law deals with the simplest case:
• The net force on a body is zero
♦ As a result, a body at rest remains at rest
♦ As a result, a body in motion remains in motion
■ The second law deals with the general situation:
• The net force on the body is not zero
♦ That is., a net force is indeed acting on the body.
We will consider both the cases: Body at rest and body in uniform motion
That is., we want to know:
A. What happens to a body at rest, if a net force is acting on it
B. What happens to a body in motion, if a net force is acting on it
Let us write the steps:
A. Body at rest
• What happens when a net force act on a body at rest?
That is., we want to know:
A. What happens to a body at rest, if a net force is acting on it
B. What happens to a body in motion, if a net force is acting on it
Let us write the steps:
A. Body at rest
• What happens when a net force act on a body at rest?
• To move means, it has a velocity
♦ This velocity may be uniform or non-uniform
• Consider the instantaneous velocity at any instant after the motion has started. Let it be v
• The initial velocity was zero
• The new velocity of v cannot be attained with out acceleration from zero value
■ So we can write:
Net force acts on a body at rest
⇒ Velocity change from zero to v
⇒ The body experienced acceleration
B. Body in uniform motion
• What happens when a net force act on a body in uniform motion?
Ans: The ‘state of uniform motion’ will change
• This change implies that velocity changes
• Consider the instantaneous velocity at any instant after the change has occurred. Let it be v2
• Let the initial velocity be v1
• The new velocity of v2 cannot be attained with out acceleration/retardation from v1
■ So we can write:
Net force acts on a body in uniform motion
⇒ Velocity change from v1 to v2
⇒ The body experienced acceleration/retardation
• So it is obvious: If a net external force act on a body, it will experience an acceleration
• For that, we will do some experiments:
Experiment 1:
1. Consider a car on a level road. Fig.5.4(a) below:
 |
| Fig.5.4 |
2. For that, we have to apply force. Let us apply force in a systematic way.
• Because, we want to take 'time' also into consideration.
3. Let a force F1 be applied from left to right
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't1' in the stop-watch will be 0
4. The car will not move at the same instant when F1 is applied. It will take some time to start moving
• If the car is big it will take a longer time
• If the car is small it will take a only a shorter time.
• Once it start moving, it's velocity will go on increasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F1 is applied
(ii) The instant when the car attains a velocity of say 2 ms-1
(we can fix any convenient value for the velocity. 2 ms-1 is only an example)
6. To measure Δt, we must carefully note down the stop-watch reading 't2' at the instant when the speedometer reading reaches 2 ms-1
• Let t2 = t
• Then Δt = (t2 - t1) = (t-0) = t s
7. So we can write:
• A force F1 is required to push a car from rest, to move it with a velocity of 2 ms-1
• The time required for this velocity change (from zero to 2 ms-1) is t s
Experiment 2:
• We repeat the experiment with a truck. See fig.5.4(b) above
• This time there is a difference:
♦ We want to push the truck from rest, to move it with the same velocity of 2 ms-1.
♦ We want to achieve this 'velocity change' within the same duration 't' that we obtained in experiment 1
Experiment 2:
• We repeat the experiment with a truck. See fig.5.4(b) above
• This time there is a difference:
♦ We want to push the truck from rest, to move it with the same velocity of 2 ms-1.
♦ We want to achieve this 'velocity change' within the same duration 't' that we obtained in experiment 1
1. Consider a truck on the same road in experiment 1
• It is in a state of rest. We want to push it manually to the right
2. For that, we have to apply force. Let us apply force in a systematic way.
• Because, we want to take 'time' also into consideration.
3. Let a force F be applied from left to right
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't1' in the stop-watch will be 0
4. The truck will not move at the same instant when F is applied. It will take some time to start moving
• Once it starts moving, it's velocity will go on increasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F is applied
(ii) The instant when the truck attains a velocity of say 2 ms-1
6. To measure Δt, we must carefully note down the stop-watch reading 't2' at the instant when the speedometer reading is 2 ms-1
• Let t2 = t
• Then Δt = (t2 - t1) = (t-0) = t s
7. But now there is a problem.
• We want both time and velocity to be the same as in experiment 1.
• That is., we want to attain the velocity 2 ms-1 in the same interval 't' obtained in experiment 1
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the truck to rest and start pushing it again
8. We do the trials until the following two conditions are satisfied
(i) The truck attains a velocity of 2 ms-1.
(ii) This velocity is attained in the same time duration 't' as in experiment 1
• The force F which satisfies the both two conditions can be noted down as F2
■ We can write:
• A force F2 is required to push the truck from rest, to move it with a velocity of 2 ms-1.
• The time required for this velocity change (from zero to 2 ms-1) is the same t s as in experiment 1
• The experiment 2 is over
Now we make a comparison between the results of the two experiments
Now we make a comparison between the results of the two experiments
• The truck obviously has greater mass than the car
■ We can write: When mass increases, greater force is required.
Let us do another set of two experiments to confirm this:
Experiment 3:
1. Consider a 'trolley carrying a mass' on a level road. See fig.5.5(a) below:
 |
| Fig.5.5 |
• It's velocity is 1.5 ms-1.
• We want to bring it to a stop
2. For that, we have to apply force towards the left. Let us apply force in a systematic way.
• Because, we want to take 'time' also into consideration.
3. Let a force F3 be applied from right to left
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't1' in the stop-watch will be 0
4. The trolley will not stop at the same instant when F3 is applied. It will take some time to stop
• If the trolley contains more mass, it will take a longer time
• If the trolley contains less mass, it will take a only a shorter time.
• Once it start to slow down, it's velocity will go on decreasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F3 is applied
(ii) The instant when the trolley attains a velocity of 0 ms-1
6. To measure Δt, we must carefully note down the stop-watch reading 't2' at the instant when the trolley comes to rest
• Let t2 = t
• Then Δt = (t2 - t1) = (t-0) = t s
7. So we can write:
• A force F3 is required to bring the trolley to rest.
• The time required for this velocity change (from 1.5 ms-1 to zero) is t s
Experiment 4:
• We repeat the experiment after increasing the mass on the trolley. Also the velocity should be the same '1.5 ms-1' as before. See. fig.5.5(b) above
• However, this time there is a difference:
♦ We want to bring the new mass to rest
♦ We want to achieve this 'velocity change' (from 1.5 to zero) within the same duration 't' that we obtained in experiment 3
• We repeat the experiment after increasing the mass on the trolley. Also the velocity should be the same '1.5 ms-1' as before. See. fig.5.5(b) above
• However, this time there is a difference:
♦ We want to bring the new mass to rest
♦ We want to achieve this 'velocity change' (from 1.5 to zero) within the same duration 't' that we obtained in experiment 3
1. Consider the 'trolley with increased mass' on the same floor as in experiment 3
• It is in a state of uniform motion.
• It's velocity is 1.5 ms-1.
• We want to bring it to a stop
2. For that, we have to apply force towards the left. Let us apply force in a systematic way.
• Because, we want to take 'time' also into consideration.
3. Let a force F be applied from right to left
• At the ‘instant when this force is applied’, turn on the stop-watch.
• So the reading 't1' in the stop-watch will be 0
4. The trolley will not stop at the same instant when F is applied. It will take some time to stop
• Once it start to slow down, it's velocity will go on decreasing
• The measurement of 'time' is important for our present experiment
5. That is., we want the interval of time Δt between the following to instances:
(i) The instant when force F3 is applied
(ii) The instant when the trolley attains a velocity of 0 ms-1
6. To measure Δt, we must carefully note down the stop-watch reading 't2' at the instant when the trolley comes to rest
• Let t2 = t
• Then Δt = (t2 - t1) = (t-0) = t s
7. But now there is a problem.
• We want both time and velocity to be the same as in experiment 1.
• That is., we want to attain the velocity 0 ms-1 in the same interval 't' obtained in experiment 3
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the trolley to 'uniform motion at 1.5 ms-1' and try to stop it again
8. We do the trials until the following two conditions are satisfied
(i) The trolley attains a velocity of 0 ms-1.
(ii) This velocity is attained in the same time duration 't' as in experiment 3
• The force F which satisfies the both two conditions can be noted down as F4
■ We can write:
• A force F4 is required to bring the trolley to rest.
• The time required for this velocity change (from 1.5 ms-1 to zero) is the same t s obtained in experiment 3
• The experiment 4 is over
Now we make a comparison between the results of the two experiments 3 and 4
Now we make a comparison between the results of the two experiments 3 and 4
• The trolley in experiment 4 obviously has greater mass than that in experiment 1
■ We can write: When mass increases, greater force is required.
Let us do one more 'set of two experiments' to confirm this. These are simple experiments:
Experiment 5:
1. Drop a small stone from the top of a building
2. Let a person standing at the foot of the building catch it
3. The following points should be noted while catching:
♦ The person must wear a pair of good quality work gloves. This is to avoid injury.
♦ The person must not lower his hands while making the catch
4. Let the force experienced by the person be F5
Experiment 6:
1. Drop a stone which is a little bigger (than the one used in experiment 5) from the top of the same building
2. Let the person standing at the foot of the building catch it
3. The following points should be noted this time also:
♦ The person must wear a pair of good quality work gloves. This is to avoid injury.
♦ The person must not lower his hands while making the catch
4. Let the force experienced by the person be F6
• Experiment 6 is over.
• Let us make a comparison between the results of experiments 5 and 6
• Let us make a comparison between the results of experiments 5 and 6
The following points may be noted:
(i) The bigger stone has greater mass
(ii) The height is same in both experiments
♦ So the 'velocity at the time of catch' will be the same in both cases
(iii) The person does not lower his hands
♦ So the 'time of application of the force' is same in both cases
■ So we completed three sets of experiments. We can now write with confidence:
• When mass decreases, lesser force is required.
In other words:
■ Force is directly proportional to mass
• symbolically we write this as: F∝ m
♦ Where 'F' represents the force and 'm' represents the mass
♦ The symbol '∝' stands for 'is proportional to'
■ Thus we found the relation between mass and force.
• In the next section we will obtain the relation between velocity and force
♦ For that, we will change the velocity while keeping mass and time constant
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