In the previous chapter, we saw Newton's Second law of motion. We also saw that Force is a product of mass and acceleration. In this section we will see some cases showing practical applications of this law.
Case 1
1. Consider a fast moving cricket ball. It is moving towards a fielder. The fielder wants to catch it.
2. Just when the ball touches his hands, he gradually withdraws his hands.
It requires some skill. Because, while withdrawing his hands, the fielder should not lose the ball.
3. If a skilled player withdraws his hands, the ball will be satisfying both the following two conditions:
• The ball will be touching the withdrawing hands all the time.
• The ball will be moving with a certain velocity.
4. When the withdrawal stops, the velocity will be zero, and the ball will be safely in his hands.
5. Now, as the ball is in touch with the withdrawing hands, it is brought to a stop gradually. That means the stopping is not instantaneous.
6. Let the velocity with which the ball approaches the fielder (the initial velocity) be u.
The final velocity of the ball is zero.
7. The change in velocity is achieved in a time interval 't'. This interval is between the following two:
• The time of first contact with the hands
• The time of stopping the withdrawal of the hands.
8. If the mass of the ball is m, then the force F required to stop the ball is given by:
F = ma = m×(v-u)⁄t = m×(0-u)⁄t = -mu⁄t (The negative sign indicates that, the force is applied in the opposite direction of the movement of the ball)
9. The time 't' is in the denominator. So if t increases, the force decreases. That means, he need to put only a little effort to stop it.
• If he does not withdraw his hands, the time in which the ball is brought to rest is less. As t is in the denominator, the force will be more. That means he will have to put more effort to stop the ball, and may cause injury to the palms.
■ The above steps can be written in short:
1. When a fielder catches a cricket ball, he gradually withdraws his hands
2. So the velocity of the ball gradually decreases to zero.
3. Gradual decrease in velocity means that, there is a time duration t in which the velocity is brought to zero. 4. This time duration in the denominator will reduce the force.
5. If hands are not withdrawn, t will be very small, thus increasing the force and there by causing injuries.
Case 2:
This is a case similar to case 1.
1. When an athlete performs the high jump event, he is made to land on a cushioned bed.
2. When he lands, the cushion compresses down, and his velocity gradually decreases to zero.
3. Gradual decrease in velocity means that, there is a time duration t in which the velocity is brought to zero.
4. This time duration in the denominator will reduce the force.
5. If cushion is not provided, t will be very small, thus increasing the force and there by causing injuries.
How much force does the table exert on the ball to bring it to rest?
Solution:
1. A ball was moving along a straight line on a long table. We wanted to make a velocity-time graph of the motion of the ball.
2. So we started a stop watch. At the instant when the stop watch is started, t1 = 0.00 s
3. Remember the ball was in motion when the stop watch was started. At the instant of starting the stop watch, the velocity of the ball = initial velocity u = 20 cm s-1. This value is obtained from the given graph. So we can write:
u = 20 cm s-1 = 0.2 m s-1
4. When the ball came to rest, it's velocity = final velocity v = 0
5. At that instant when the ball came to rest, the reading in the stop watch showed t2 = 10.00 s
6. So time of travel during the experiment = t = t2 - t1 = 10.00 - 0.00 = 10 s
7. So acceleration = (v-u)⁄t = (0-0.2)⁄10 = -0.02 m s-2. The negative sign shows that it is deceleration.
8. When the ball moves on the table, the table exerts a pulling force F on the ball due to friction. That is why the ball is decelerated to zero velocity.
9. We have F = ma
m = 20 g = 20⁄1000 = 0.02 kg
10. So F = 0.02 × (-0.02) = - 0.0004 N. (The negative sign indicates that, the frictional force is applied in the opposite direction of motion of the ball)
Note: To get the force in N, mass must be in kg, and acceleration must be in m s-2. This we saw at the end of the previous section. So velocity was converted from cm s-1 to m s-1 in step (3). Also mass was converted from grams to kilograms in step (9)
Solved example 2.6
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
Solution:
Part (i). To find the acceleration:
1. u = 0, s = 400 m, t = 20 s. a = ?
2. We can use the second equation of motion. Because it connects the above four quantities.
3. s = ut + 1⁄2 at2 ⇒ 400 = 0 × 20 + 1⁄2 × a × 202 ⇒ 400 = 200a a = 2 m s-2.
Part (ii). To find the force:
1. If an object is moving with an acceleration, a force F must be acting on it.
2. This force is given by: F = ma
3. We have mass = m = 7 tonnes = 7 × 1000 = 7000 kg
4. Thus F = 7000 × 2 = 14000 N
Solved example 2.7
A stone of 1 kg is thrown with a velocity of 20 m s-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Solution:
1. When a stone is thrown across the frozen surface of a lake, it will slide on the surface and travel some distance. This is because of the lesser friction offered by the ice.
2. Given: u = 20 m s-1, v = 0 (∵ the stone comes to rest), s = 50 m
3. The velocity is reduced to zero. That means there is deceleration, which is 'negative acceleration'. So we have to find this negative acceleration first.
4. We can use the third equation of motion. Because it connects all the quantities in (2)
v2 = u2 + 2as ⇒ 02 = 202 + 2 × a × 50 ⇒ 0 = 400 + 100a ⇒ a = -4 m s-2.
5. So the stone moved with an acceleration a = -4 m s-2. This 'slowing down' was due to a force F acting on the stone.
6. This F is nothing but the 'frictional resistance' offered by the ice. In other words, it is the 'force of friction between the stone and the ice'.
7. This F causes an acceleration of -4 m s-2 on the stone of mass 1 kg.
8. So using the equation F = ma, we get:
F = 1 × (-4) = -4 N
Solved example 2.8
An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s-2 ?
Solution:
1. The vehicle has a mass of 1500 kg. This mass has to be stopped using the 'frictional resistance' force F between the vehicle and the road. No brakes are to be applied.
2. If the vehicle is to be stopped, there must be negative acceleration. It is given as -1.7 m s-2.
3. So F must produce this negative acceleration. We can use F = ma
We get: F = 1500 × (-1.7) = -2550 N (The negative sign indicates that, the frictional force is applied in the opposite direction of motion of the vehicle)
Solved example 2.9
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Solution:
1. We are applying a horizontal force of 200 N.
2. The frictional force F will also be horizontal. Because, it acts parallel to the floor. We want the magnitude and direction of this F.
3. It is given that the cabinet must move with a constant velocity. That means, there will be no acceleration. Because, if there is acceleration or deceleration, the velocity will change.
4. If there is no acceleration or deceleration, it is obvious that no force is acting on the cabinet. Because we know that F = ma
5. That means, the applied force of 200 N is cancelled by the frictional force F
6. That means 200 N and F are equal and opposite
7. That means F = -200 N
8. So we can write: Frictional force exerted on the cabinet is 200 N in the direction opposite to the applied 200 N
In the next section, we will see Newton's Third Law of motion.
Case 1
1. Consider a fast moving cricket ball. It is moving towards a fielder. The fielder wants to catch it.
2. Just when the ball touches his hands, he gradually withdraws his hands.
It requires some skill. Because, while withdrawing his hands, the fielder should not lose the ball.
3. If a skilled player withdraws his hands, the ball will be satisfying both the following two conditions:
• The ball will be touching the withdrawing hands all the time.
• The ball will be moving with a certain velocity.
4. When the withdrawal stops, the velocity will be zero, and the ball will be safely in his hands.
5. Now, as the ball is in touch with the withdrawing hands, it is brought to a stop gradually. That means the stopping is not instantaneous.
6. Let the velocity with which the ball approaches the fielder (the initial velocity) be u.
The final velocity of the ball is zero.
7. The change in velocity is achieved in a time interval 't'. This interval is between the following two:
• The time of first contact with the hands
• The time of stopping the withdrawal of the hands.
8. If the mass of the ball is m, then the force F required to stop the ball is given by:
F = ma = m×(v-u)⁄t = m×(0-u)⁄t = -mu⁄t (The negative sign indicates that, the force is applied in the opposite direction of the movement of the ball)
9. The time 't' is in the denominator. So if t increases, the force decreases. That means, he need to put only a little effort to stop it.
• If he does not withdraw his hands, the time in which the ball is brought to rest is less. As t is in the denominator, the force will be more. That means he will have to put more effort to stop the ball, and may cause injury to the palms.
■ The above steps can be written in short:
1. When a fielder catches a cricket ball, he gradually withdraws his hands
2. So the velocity of the ball gradually decreases to zero.
3. Gradual decrease in velocity means that, there is a time duration t in which the velocity is brought to zero. 4. This time duration in the denominator will reduce the force.
5. If hands are not withdrawn, t will be very small, thus increasing the force and there by causing injuries.
Case 2:
This is a case similar to case 1.
1. When an athlete performs the high jump event, he is made to land on a cushioned bed.
2. When he lands, the cushion compresses down, and his velocity gradually decreases to zero.
3. Gradual decrease in velocity means that, there is a time duration t in which the velocity is brought to zero.
4. This time duration in the denominator will reduce the force.
5. If cushion is not provided, t will be very small, thus increasing the force and there by causing injuries.
Now we will see some solved examples
Solved example 2.1
A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object’s velocity from 3 m s-1 to 7 m s-1. Find the magnitude of the applied force. Now, if the force was applied for a duration of 5 s, what would be the final velocity of the object?
Solution:
Part (i):
1. The object was moving with a velocity of 3m s-1. So u = 3 m s-1.
2. A force F acts on that object. As a result, the velocity increases to 7 m s-1. So v = 7 m s-1.
3. If there is an increase in velocity, there must be an acceleration. [The opposite is also true: If there is a decrease in velocity, there must be a deceleration. Deceleration is nothing but acceleration in the opposite direction]
4. If we can find this acceleration a, we can obtain the force. Because Force = mass × acceleration
So our first aim is to find a.
5. We have: a = (v-u)⁄t = (7-3)⁄2 = 4⁄2 = 2 m s-2. [In this step t is taken as 2 s. Because it is given that the force acts for 2 s. That means the change of velocity from u to v takes place in a duration of 2 s]
6. Given that mass = m = 5 kg
7. So F = ma = 5 × 2 = 10 N
Note: To get the force in N, mass must be in kg, and acceleration must be in m s-2. This we saw at the end of the previous section.
Once we understand the above basics, we need not write such descriptive steps. The steps can be summarized as follows:
1. The object was moving with a velocity of 3m s-1. So u = 3 m s-1.
2. A force F acts on that object. As a result, the velocity increases to 7 m s-1. So v = 7 m s-1.
3. If there is an increase in velocity, there must be an acceleration. [The opposite is also true: If there is a decrease in velocity, there must be a deceleration. Deceleration is nothing but acceleration in the opposite direction]
4. If we can find this acceleration a, we can obtain the force. Because Force = mass × acceleration
So our first aim is to find a.
5. We have: a = (v-u)⁄t = (7-3)⁄2 = 4⁄2 = 2 m s-2. [In this step t is taken as 2 s. Because it is given that the force acts for 2 s. That means the change of velocity from u to v takes place in a duration of 2 s]
6. Given that mass = m = 5 kg
7. So F = ma = 5 × 2 = 10 N
Note: To get the force in N, mass must be in kg, and acceleration must be in m s-2. This we saw at the end of the previous section.
Once we understand the above basics, we need not write such descriptive steps. The steps can be summarized as follows:
• m = 5 kg, t = 2 s, u = 3m s-1, v = 7m s-1. F = ?
• We have: F = ma = m × (v-u)⁄t = 5 × (7-3)⁄2 = 5 × 4⁄2 = 10 N
Part (ii):
This is similar to part (i). But we are asked the final velocity when the duration changes to 5 s. Let us analyse:
1. The object was moving with a velocity of 3m s-1. So u = 3 m s-1.
2. A force F acts on that object. As a result, the velocity changes to v. We have to find this v. Force F was calculated in part (i). We got F = 10 N
3. If there is a change in velocity, there must be an acceleration or deceleration. [If the final velocity is greater, it is acceleration. If the final velocity is lesser, it is deceleration. In this problem, we do not know what the final velocity is. So we do not know whether it is acceleration or deceleration]
4. If we can find the acceleration, we can find the final velocity. Because a = (v-u)⁄t.
So our first aim is to find a
5. We have F = ma ⇒ 10 = 5a ⇒ a = 2 m s-2.
6. Substituting this value of a in (4) we get:
2 = (v-3)⁄5 ⇒ 10 = v-3 ⇒ v = 13 m s-1.
Once we understand the above basics, we need not write such descriptive steps. The steps can be summarized as follows:
Part (ii):
This is similar to part (i). But we are asked the final velocity when the duration changes to 5 s. Let us analyse:
1. The object was moving with a velocity of 3m s-1. So u = 3 m s-1.
2. A force F acts on that object. As a result, the velocity changes to v. We have to find this v. Force F was calculated in part (i). We got F = 10 N
3. If there is a change in velocity, there must be an acceleration or deceleration. [If the final velocity is greater, it is acceleration. If the final velocity is lesser, it is deceleration. In this problem, we do not know what the final velocity is. So we do not know whether it is acceleration or deceleration]
4. If we can find the acceleration, we can find the final velocity. Because a = (v-u)⁄t.
So our first aim is to find a
5. We have F = ma ⇒ 10 = 5a ⇒ a = 2 m s-2.
6. Substituting this value of a in (4) we get:
2 = (v-3)⁄5 ⇒ 10 = v-3 ⇒ v = 13 m s-1.
Once we understand the above basics, we need not write such descriptive steps. The steps can be summarized as follows:
• m = 5 kg, t = 5 s, u = 3m s-1, F = 10 N. v = ?
• We have: F = ma = m × (v-u)⁄t
⇒ 10 = 5 × (v-3)⁄5 = (v-3) ⇒ 10 = v-3
• So v = 10+3 = 13 m s-1.
Solved example 2.2
Solved example 2.2
Which would require a greater force – accelerating a 2 kg mass at 5 m s-2 or a 4 kg mass at 2 m s-2 ?
Solution:
Case 1:
1. A mass of 2 kg is travelling at a velocity of u. [This u can be zero or any other value. If it is zero, it would mean that, the body is initially at rest. And that the body is accelerated from rest]
2. It has to be given an acceleration of 5 m s-2.
3. How much force is required to give that much acceleration?
This is a simple problem. We know that any force would produce an acceleration. And that force is given by F = ma = 2 × 5 = 10 N.
Once we understand the above basics, we need not write such descriptive steps. The steps can be summarized as follows:
1. A mass of 2 kg is travelling at a velocity of u. [This u can be zero or any other value. If it is zero, it would mean that, the body is initially at rest. And that the body is accelerated from rest]
2. It has to be given an acceleration of 5 m s-2.
3. How much force is required to give that much acceleration?
This is a simple problem. We know that any force would produce an acceleration. And that force is given by F = ma = 2 × 5 = 10 N.
Once we understand the above basics, we need not write such descriptive steps. The steps can be summarized as follows:
• m = 2 kg, a = 5 m s-2.
• We have: F = ma = 2 × 5 = 10 N
Case 2:
• m = 4 kg, a = 2 m s-2.
• We have: F = ma = 4 × 2 = 8 N
■ So case 1 requires a greater force.
Solved example 2.3
Solved example 2.3
A motorcar is moving with a velocity of 108 kmph and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg.
Solution:
1. Given that the car is moving with a velocity of 108 kmph. So initial velocity u = 108 kmph = (108 × 1000)⁄(1 × 3600) = 30 m s-1.
2. When the brakes begin to take action, a stopping force F is acting on the car. This F acts for 4 s. So t = 4 s
3. As a result of the application of F, the car comes to a stop, That is final velocity = 0. So v = 0
4. Mass of the car = m = 1000 kg.
5. We have to find the stopping force F
We have F = ma = m × (v-u)⁄t = 1000 × (0-30)⁄4 = 1000 × -30⁄4 = -7500 N (The negative sign indicates that, the force is applied in the opposite direction of the movement of the car)
Note: To get the force in N, mass must be in kg, and acceleration must be in m s-2. This we saw at the end of the previous section.
• To get the acceleration (v-u)⁄t in m s-2, the velocities must be in m s-1.
• That is why we converted the 108 kmph velocity into 30 m s-1 velocity in step (1)
Solved example 2.4
A force of 5 N gives a mass m1, an acceleration of 10 m s-2 and a mass m2, an acceleration of 20 m s-2. What acceleration would it give if both the masses were tied together?
Solution:
1. We have F = ma ⇒ 5 = m1 × 10 ⇒ m1 = 0.5 kg
2. Also 5 = m2 × 20 ⇒ m2 = 0.25 kg
3. When they are tied together, the total mass becomes (m1 + m2) = 0.5 + 0.25 = 0.75 kg
4. So F = ma ⇒ a = F⁄m = 5⁄0.75 = 6.67 m s-2.
Solved example 2.5
The velocity-time graph of a ball of mass 20 g moving along a straight line on a long table is given in fig.2.7 below:
Solution:
1. A ball was moving along a straight line on a long table. We wanted to make a velocity-time graph of the motion of the ball.
2. So we started a stop watch. At the instant when the stop watch is started, t1 = 0.00 s
3. Remember the ball was in motion when the stop watch was started. At the instant of starting the stop watch, the velocity of the ball = initial velocity u = 20 cm s-1. This value is obtained from the given graph. So we can write:
u = 20 cm s-1 = 0.2 m s-1
4. When the ball came to rest, it's velocity = final velocity v = 0
5. At that instant when the ball came to rest, the reading in the stop watch showed t2 = 10.00 s
6. So time of travel during the experiment = t = t2 - t1 = 10.00 - 0.00 = 10 s
7. So acceleration = (v-u)⁄t = (0-0.2)⁄10 = -0.02 m s-2. The negative sign shows that it is deceleration.
8. When the ball moves on the table, the table exerts a pulling force F on the ball due to friction. That is why the ball is decelerated to zero velocity.
9. We have F = ma
m = 20 g = 20⁄1000 = 0.02 kg
10. So F = 0.02 × (-0.02) = - 0.0004 N. (The negative sign indicates that, the frictional force is applied in the opposite direction of motion of the ball)
Note: To get the force in N, mass must be in kg, and acceleration must be in m s-2. This we saw at the end of the previous section. So velocity was converted from cm s-1 to m s-1 in step (3). Also mass was converted from grams to kilograms in step (9)
Solved example 2.6
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
Solution:
Part (i). To find the acceleration:
1. u = 0, s = 400 m, t = 20 s. a = ?
2. We can use the second equation of motion. Because it connects the above four quantities.
3. s = ut + 1⁄2 at2 ⇒ 400 = 0 × 20 + 1⁄2 × a × 202 ⇒ 400 = 200a a = 2 m s-2.
Part (ii). To find the force:
1. If an object is moving with an acceleration, a force F must be acting on it.
2. This force is given by: F = ma
3. We have mass = m = 7 tonnes = 7 × 1000 = 7000 kg
4. Thus F = 7000 × 2 = 14000 N
Solved example 2.7
A stone of 1 kg is thrown with a velocity of 20 m s-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Solution:
1. When a stone is thrown across the frozen surface of a lake, it will slide on the surface and travel some distance. This is because of the lesser friction offered by the ice.
2. Given: u = 20 m s-1, v = 0 (∵ the stone comes to rest), s = 50 m
3. The velocity is reduced to zero. That means there is deceleration, which is 'negative acceleration'. So we have to find this negative acceleration first.
4. We can use the third equation of motion. Because it connects all the quantities in (2)
v2 = u2 + 2as ⇒ 02 = 202 + 2 × a × 50 ⇒ 0 = 400 + 100a ⇒ a = -4 m s-2.
5. So the stone moved with an acceleration a = -4 m s-2. This 'slowing down' was due to a force F acting on the stone.
6. This F is nothing but the 'frictional resistance' offered by the ice. In other words, it is the 'force of friction between the stone and the ice'.
7. This F causes an acceleration of -4 m s-2 on the stone of mass 1 kg.
8. So using the equation F = ma, we get:
F = 1 × (-4) = -4 N
Solved example 2.8
An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s-2 ?
Solution:
1. The vehicle has a mass of 1500 kg. This mass has to be stopped using the 'frictional resistance' force F between the vehicle and the road. No brakes are to be applied.
2. If the vehicle is to be stopped, there must be negative acceleration. It is given as -1.7 m s-2.
3. So F must produce this negative acceleration. We can use F = ma
We get: F = 1500 × (-1.7) = -2550 N (The negative sign indicates that, the frictional force is applied in the opposite direction of motion of the vehicle)
Solved example 2.9
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Solution:
1. We are applying a horizontal force of 200 N.
2. The frictional force F will also be horizontal. Because, it acts parallel to the floor. We want the magnitude and direction of this F.
3. It is given that the cabinet must move with a constant velocity. That means, there will be no acceleration. Because, if there is acceleration or deceleration, the velocity will change.
4. If there is no acceleration or deceleration, it is obvious that no force is acting on the cabinet. Because we know that F = ma
5. That means, the applied force of 200 N is cancelled by the frictional force F
6. That means 200 N and F are equal and opposite
7. That means F = -200 N
8. So we can write: Frictional force exerted on the cabinet is 200 N in the direction opposite to the applied 200 N
In the next section, we will see Newton's Third Law of motion.
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