In the previous section we obtained the relation between force and 'change in velocity (v2-v1)'.
• We saw that force is directly proportional to '(v2-v1)'.
• In this section, we will see the relation between time and velocity.
■ In the earlier section, we obtained F ∝ m by:
• Changing mass (m)
• Keeping velocity (v) and time (t) constant
■ In the previous section, we obtained F ∝ (v2-v1) by:
• Changing velocities
• Keeping velocity (v) and time (t) constant
■ Now, we will obtain the relation between F and t by:
• Changing t
• Keeping m and v constant
We will do a series of experiments here also:
Experiment 13:
1. Consider a car on a level road. See fig.5.8(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 F13 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 F13 is applied. It will take some time to start moving
• 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 F13 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 is 2 ms-1
• Let t2 = 12
• Then Δt = (t2 - t1) = (5-0) = 12 s
7. So we can write:
• A force F13 is required to push a car from rest and to move it with a velocity of 2 ms-1
• We saw that force is directly proportional to '(v2-v1)'.
• In this section, we will see the relation between time and velocity.
■ In the earlier section, we obtained F ∝ m by:
• Changing mass (m)
• Keeping velocity (v) and time (t) constant
■ In the previous section, we obtained F ∝ (v2-v1) by:
• Changing velocities
• Keeping velocity (v) and time (t) constant
■ Now, we will obtain the relation between F and t by:
• Changing t
• Keeping m and v constant
We will do a series of experiments here also:
Experiment 13:
1. Consider a car on a level road. See fig.5.8(a) below:
 |
| Fig.5.8 |
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 F13 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 F13 is applied. It will take some time to start moving
• 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 F13 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 is 2 ms-1
• Let t2 = 12
• Then Δt = (t2 - t1) = (5-0) = 12 s
7. So we can write:
• A force F13 is required to push a car from rest and to move it with a velocity of 2 ms-1
• The time required for this velocity change (from zero to 2 ms-1) is 12 s
Experiment 14:
1. Consider the same car on the same road. See fig.5.8(b)
• 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 car will not move at the same instant when F is applied. It will take some time to start moving
• 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 F is applied
(ii) The instant when the car attains the same velocity in experiment 13. That is 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
7. But now there is a problem.
• We want both mass and velocity to be the same as in experiment 13.
• It is the 'time' that we want to change
• That is., we want to attain the velocity 2 ms-1 'in a shorter time' than the time in experiment 13
Let us fix it as 8 s
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the car to rest and start pushing it again
8. We do the trials until the following two conditions are satisfied
(i) The car attains a velocity of 2 ms-1.
(ii) This velocity is attained in a time duration of 8 s
• The force F which satisfies the both two conditions can be noted down as F14
■ We can write:
• A force F14 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 8 s
■ We will find that F13 < F14.
The following points may be noted:
(i) We wanted to obtain the same effect as in experiment 13
(ii) We wanted to obtain the same effect in lesser time
(iii) So obviously the force required will be more
■ We can write: When 'time' decreases, greater force is required.
Let us do another set of two experiments to confirm this:
Experiment 15:
1. Consider a 'trolley carrying a mass' on a level road. See fig.5.9(a) below
• It is in a state of uniform motion towards the right.
• It's velocity is 2 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 F15 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 F15 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 F15 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 = 5
• Then Δt = (t2 - t1) = (3-0) = 5 s
7. So we can write:
• A force F15 is required to bring the trolley to rest.
• The time required for this velocity change (from 2 ms-1 to zero) is 5 s
• We will write the steps:
1. Consider the 'trolley with the same mass' on the same floor as in experiment 15
• It is in a state of uniform motion.
• It's velocity is 2 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 two instances:
(i) The instant when force F 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
7. But now there is a problem.
• We want both mass and velocity to be the same as in experiment 15.
• Also, we want to attain the velocity 0 ms-1 in a lesser interval '3 s'
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the trolley to 'uniform motion at 2 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 3 s
• The force F which satisfies the both two conditions can be noted down as F16
■ We can write:
• A force F16 is required to bring the trolley to rest.
• The time required for this velocity change (from 2 ms-1 to zero) is 3 s, which is less than the 5 s obtained in experiment 15
■ We will find that F15 < F16
The following points may be noted:
(i) We wanted to obtain the same effect as in experiment 15
(ii) We wanted to obtain the same effect in lesser time
(iii) So obviously, the force required will be more
■ We can write: When 'time' decreases, greater force is required.
Let us do one more 'set of two experiments' to confirm this. These are simple experiments:
Experiment 17:
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 F17
Experiment 18:
1. Drop the same stone which was used in experiment 17
• Also, the drop must be made from the same building, and from the same height in experiment 17
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 lower his hands gradually while making the catch
♦ This method of catching will increase the 'time of stopping'
4. Let the force experienced by the person be F18.
■ It will be found that F18 < F17
• The following points may be noted:
(i) The stone was dropped from the same height in both experiments
♦ So the final velocity is the same in both the cases
(ii) The same stone was dropped in both the experiments
♦ So the mass remains the same
(iii) The person lower his hands gradually in experiment 18
♦ So more time was allowed to stop the stone
■ We can say
• We wanted to obtain the same effect as in experiment 17
• Greater time was allowed to obtain the same effect
• So the force required will be less
• When 'time' decreases, greater force is required
• When 'time' increases, lesser force is required
In other words:
■ Force is inversely proportional to 'time'
• symbolically we write this as: F∝ 1⁄t
• The symbol '∝' stands for 'is proportional to'
• In the next section we will combine all the information together
Experiment 14:
1. Consider the same car on the same road. See fig.5.8(b)
• 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 car will not move at the same instant when F is applied. It will take some time to start moving
• 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 F is applied
(ii) The instant when the car attains the same velocity in experiment 13. That is 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
7. But now there is a problem.
• We want both mass and velocity to be the same as in experiment 13.
• It is the 'time' that we want to change
• That is., we want to attain the velocity 2 ms-1 'in a shorter time' than the time in experiment 13
Let us fix it as 8 s
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the car to rest and start pushing it again
8. We do the trials until the following two conditions are satisfied
(i) The car attains a velocity of 2 ms-1.
(ii) This velocity is attained in a time duration of 8 s
• The force F which satisfies the both two conditions can be noted down as F14
■ We can write:
• A force F14 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 8 s
• The experiment 14 is over
Now we make a comparison between the results of the two experiments
The following points may be noted:
(i) We wanted to obtain the same effect as in experiment 13
(ii) We wanted to obtain the same effect in lesser time
(iii) So obviously the force required will be more
■ We can write: When 'time' decreases, greater force is required.
Let us do another set of two experiments to confirm this:
Experiment 15:
1. Consider a 'trolley carrying a mass' on a level road. See fig.5.9(a) below
 |
| Fig.5.9 |
• It's velocity is 2 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 F15 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 F15 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 F15 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 = 5
• Then Δt = (t2 - t1) = (3-0) = 5 s
7. So we can write:
• A force F15 is required to bring the trolley to rest.
• The time required for this velocity change (from 2 ms-1 to zero) is 5 s
Experiment 16:
• We repeat the experiment with the same trolley used in the previous experiment 15.
• The mass contained inside it should not change.
• The following points should be noted:
♦ The trolley must be moving with the same velocity (2 ms-1) as in experiment 15
♦ We want to bring the trolley to rest
♦ We want to achieve this 'velocity change' (from 2 to zero) within a shorter time than in experiment 15. Let us fix it at 3 s
• We repeat the experiment with the same trolley used in the previous experiment 15.
• The mass contained inside it should not change.
• The following points should be noted:
♦ The trolley must be moving with the same velocity (2 ms-1) as in experiment 15
♦ We want to bring the trolley to rest
♦ We want to achieve this 'velocity change' (from 2 to zero) within a shorter time than in experiment 15. Let us fix it at 3 s
1. Consider the 'trolley with the same mass' on the same floor as in experiment 15
• It is in a state of uniform motion.
• It's velocity is 2 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 two instances:
(i) The instant when force F 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
7. But now there is a problem.
• We want both mass and velocity to be the same as in experiment 15.
• Also, we want to attain the velocity 0 ms-1 in a lesser interval '3 s'
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the trolley to 'uniform motion at 2 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 3 s
• The force F which satisfies the both two conditions can be noted down as F16
■ We can write:
• A force F16 is required to bring the trolley to rest.
• The time required for this velocity change (from 2 ms-1 to zero) is 3 s, which is less than the 5 s obtained in experiment 15
• The experiment 16 is over
Now we make a comparison between the results of the two experiments
The following points may be noted:
(i) We wanted to obtain the same effect as in experiment 15
(ii) We wanted to obtain the same effect in lesser time
(iii) So obviously, the force required will be more
■ We can write: When 'time' decreases, greater force is required.
Let us do one more 'set of two experiments' to confirm this. These are simple experiments:
Experiment 17:
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 F17
Experiment 18:
1. Drop the same stone which was used in experiment 17
• Also, the drop must be made from the same building, and from the same height in experiment 17
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 lower his hands gradually while making the catch
♦ This method of catching will increase the 'time of stopping'
4. Let the force experienced by the person be F18.
• Let us make a comparison between the results of experiments 17 and 18
• The following points may be noted:
(i) The stone was dropped from the same height in both experiments
♦ So the final velocity is the same in both the cases
(ii) The same stone was dropped in both the experiments
♦ So the mass remains the same
(iii) The person lower his hands gradually in experiment 18
♦ So more time was allowed to stop the stone
■ We can say
• We wanted to obtain the same effect as in experiment 17
• Greater time was allowed to obtain the same effect
• So the force required will be less
■ So we completed three sets of experiments. We can now write with confidence:
• When 'time' increases, lesser force is required
In other words:
■ Force is inversely proportional to 'time'
• symbolically we write this as: F∝ 1⁄t
• The symbol '∝' stands for 'is proportional to'
• Thus we found the relation between 'time' and force.
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