In the previous section we obtained the relation between force and mass. We saw that force is directly proportional to mass. In this section, we will see the relation between force and velocity.
■ We obtained F ∝ m by:
• Changing mass (m)
• Keeping velocity (v) and time (t) constant
■ Now, we will obtain the relation between F and v by:
• Changing v
• Keeping m and t constant
We will do a series of experiments here also:
Experiment 7:
1. Consider a car on a level road. See fig.5.6(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 F7 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
• 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 F7 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 F7 is required to push a car from rest and to move it with a velocity of 2 ms-1
■ We obtained F ∝ m by:
• Changing mass (m)
• Keeping velocity (v) and time (t) constant
■ Now, we will obtain the relation between F and v by:
• Changing v
• Keeping m and t constant
We will do a series of experiments here also:
Experiment 7:
1. Consider a car on a level road. See fig.5.6(a) below:
 |
| Fig.5.6 |
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 F7 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
• 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 F7 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 F7 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 t s
Experiment 8:
1. Consider the same car on the same road. See fig.5.6(b) above
• 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 a higher velocity than in experiment 7. Say 3 ms-1
(we can fix any convenient value greater than 2 which we used in the previous experiment. 3 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 3 ms-1
• Let t2 = t
• Then Δt = (t2 - t1) = (t-0) = t s
7. But now there is a problem.
• We want both mass and time to be the same as in experiment 7.
• That is., we want to attain the velocity 3 ms-1 in the same interval 't' obtained in experiment 7
• 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 3 ms-1.
(ii) This velocity is attained in the same time duration 't' as in experiment 7
• The force F which satisfies the both two conditions can be noted down as F8
■ We can write:
• A force F8 is required to push a car from rest, to move it with a velocity of 3 ms-1.
• The time required for this velocity change (from zero to 3 ms-1) is the same t s as in experiment 7
■ We will find that F7 < F8.
The following points may be noted:
(i) In experiment 7, v1 = 0 and v2 = 2
• So change in velocity achieved = (v2 - v1) = (2-0) = 2 ms-1
(ii) In experiment 8, v1 = 0 and v2 = 3
• So change in velocity achieved = (v2 - v1) = (3-0) = 3 ms-1
(iii) The 'change in velocity' is obviously greater in experiment 8
■ We can write: When 'change in velocity' increases, greater force is required.
Let us do another set of two experiments to confirm this:
Experiment 9:
1. Consider a 'trolley carrying a mass' on a level road. See fig.4.7(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 F9 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 F9 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 F9 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 F9 is required to bring the trolley to rest.
• The time required for this velocity change (from 2 ms-1 to zero) is t s
• We will write the steps:
1. Consider the 'trolley with the same mass' on the same floor as in experiment 9
• It is in a state of uniform motion.
• It's velocity is 3 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 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
• Let t2 = t
• Then Δt = (t2 - t1) = (t-0) = t s
7. But now there is a problem.
• We want both mass and velocity to be the same as in experiment 9
• That is., we want to attain the velocity 0 ms-1 in the same interval 't' obtained in experiment 9
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the trolley to 'uniform motion at 3 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 9
• The force F which satisfies the both two conditions can be noted down as F10
■ We can write:
• A force F10 is required to bring the trolley to rest.
• The time required for this velocity change (from 3 ms-1 to zero) is the same t s obtained in experiment 9
■ We will find that F9 < F10
The following points may be noted:
(i) In experiment 9, v1 = 2 and v2 = 0
• So change in velocity achieved = (v2 - v1) = (0-2) = -2 ms-1.
(ii) In experiment 8, v1 = 3 and v2 = 0
• So change in velocity achieved = (v2 - v1) = (0-3) = -3 ms-1.
(iii) The magnitude of the 'change in velocity' is obviously greater in experiment 10
■ We can write: When 'change in velocity' increases, greater force is required.
Let us do one more 'set of two experiments' to confirm this. These are simple experiments:
Experiment 11:
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 F11
Experiment 12:
1. Drop the same stone which was used in experiment 11
• But this time, the drop must be made from a taller building than in experiment 11
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 F12.
■ We will find that F11 < F12
• The following points may be noted:
(i) The stone dropped from a greater height would have attained a greater velocity (v2)
♦ The initial velocity (v1) is zero in both cases
♦ So the change in velocity (v2-v1) will be more in experiment 12
(ii) The same stone was dropped in both the experiments
♦ So the mass remains the same
(iii) The person does not lower his hands
♦ So the 'time of application of the force' is same in both cases
• When 'change in velocity (v2-v1)' increases, greater force is required
• When 'change in velocity (v2-v1)' decreases, lesser force is required
In other words:
■ Force is directly proportional to 'change in velocity (v2-v1)'
• Symbolically, we write this as: F∝ (v2-v1)
• We obtained it by changing the 'velocity' while keeping mass and time constant
• In the next section we will obtain the relation between time and force
• For that, we will change the time while keeping mass and velocity constant
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Experiment 8:
1. Consider the same car on the same road. See fig.5.6(b) above
• 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 a higher velocity than in experiment 7. Say 3 ms-1
(we can fix any convenient value greater than 2 which we used in the previous experiment. 3 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 3 ms-1
• Let t2 = t
• Then Δt = (t2 - t1) = (t-0) = t s
7. But now there is a problem.
• We want both mass and time to be the same as in experiment 7.
• That is., we want to attain the velocity 3 ms-1 in the same interval 't' obtained in experiment 7
• 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 3 ms-1.
(ii) This velocity is attained in the same time duration 't' as in experiment 7
• The force F which satisfies the both two conditions can be noted down as F8
■ We can write:
• A force F8 is required to push a car from rest, to move it with a velocity of 3 ms-1.
• The time required for this velocity change (from zero to 3 ms-1) is the same t s as in experiment 7
• The experiment 8 is over
Now we make a comparison between the results of the two experiments
The following points may be noted:
(i) In experiment 7, v1 = 0 and v2 = 2
• So change in velocity achieved = (v2 - v1) = (2-0) = 2 ms-1
(ii) In experiment 8, v1 = 0 and v2 = 3
• So change in velocity achieved = (v2 - v1) = (3-0) = 3 ms-1
(iii) The 'change in velocity' is obviously greater in experiment 8
■ We can write: When 'change in velocity' increases, greater force is required.
Let us do another set of two experiments to confirm this:
Experiment 9:
1. Consider a 'trolley carrying a mass' on a level road. See fig.4.7(a) below:
 |
| Fig.5.7 |
• 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 F9 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 F9 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 F9 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 F9 is required to bring the trolley to rest.
• The time required for this velocity change (from 2 ms-1 to zero) is t s
Experiment 10:
• We repeat the experiment with the same trolley used in the previous experiment 9. See fig.4.7(b)
• The mass contained inside it should not change.
• But the velocity should be different. Let the trolley be moving with a uniform velocity of 3 ms-1.
• The following points should be noted:
♦ We want to bring the trolley to rest
♦ We want to achieve this 'velocity change' (from 3 to zero) within the same duration 't' that we obtained in experiment 9
• We repeat the experiment with the same trolley used in the previous experiment 9. See fig.4.7(b)
• The mass contained inside it should not change.
• But the velocity should be different. Let the trolley be moving with a uniform velocity of 3 ms-1.
• The following points should be noted:
♦ We want to bring the trolley to rest
♦ We want to achieve this 'velocity change' (from 3 to zero) within the same duration 't' that we obtained in experiment 9
1. Consider the 'trolley with the same mass' on the same floor as in experiment 9
• It is in a state of uniform motion.
• It's velocity is 3 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 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
• Let t2 = t
• Then Δt = (t2 - t1) = (t-0) = t s
7. But now there is a problem.
• We want both mass and velocity to be the same as in experiment 9
• That is., we want to attain the velocity 0 ms-1 in the same interval 't' obtained in experiment 9
• This may not be possible in just one trial.
• So we do several trials. That is., we bring the trolley to 'uniform motion at 3 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 9
• The force F which satisfies the both two conditions can be noted down as F10
■ We can write:
• A force F10 is required to bring the trolley to rest.
• The time required for this velocity change (from 3 ms-1 to zero) is the same t s obtained in experiment 9
• The experiment 10 is over
Now we make a comparison between the results of the two experiments
The following points may be noted:
(i) In experiment 9, v1 = 2 and v2 = 0
• So change in velocity achieved = (v2 - v1) = (0-2) = -2 ms-1.
(ii) In experiment 8, v1 = 3 and v2 = 0
• So change in velocity achieved = (v2 - v1) = (0-3) = -3 ms-1.
(iii) The magnitude of the 'change in velocity' is obviously greater in experiment 10
■ We can write: When 'change in velocity' increases, greater force is required.
Let us do one more 'set of two experiments' to confirm this. These are simple experiments:
Experiment 11:
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 F11
Experiment 12:
1. Drop the same stone which was used in experiment 11
• But this time, the drop must be made from a taller building than in experiment 11
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 F12.
• Let us make a comparison between the results of experiments 11 and 12
• The following points may be noted:
(i) The stone dropped from a greater height would have attained a greater velocity (v2)
♦ The initial velocity (v1) is zero in both cases
♦ So the change in velocity (v2-v1) will be more in experiment 12
(ii) The same stone was dropped in both the experiments
♦ So the mass remains the same
(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 'change in velocity (v2-v1)' decreases, lesser force is required
In other words:
■ Force is directly proportional to 'change in velocity (v2-v1)'
• Symbolically, we write this as: F∝ (v2-v1)
■ Thus we found the relation between 'change in velocity' and force.
• In the next section we will obtain the relation between time and force
• For that, we will change the time while keeping mass and velocity constant
PREVIOUS CONTENTS NEXT
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