In the previous section, we saw different properties of scalar products. In this section, we will see work done by a force and also some basics about kinetic energy.
We have seen some basics about kinetic energy in our high school classes. The following two links will give those details
High school physics - Kinetic energy lesson 1
High school physics - Kinetic energy lesson 2
It is recommended that the reader get a thorough understanding of those lessons before taking up our present discussion
We have seen some basics about kinetic energy in our high school classes. The following two links will give those details
High school physics - Kinetic energy lesson 1
High school physics - Kinetic energy lesson 2
It is recommended that the reader get a thorough understanding of those lessons before taking up our present discussion
We will write our present discussion on kinetic energy in steps:
1. In chapter 3, we saw the third equation of motion:
$\mathbf\small{v^2-v_0^2=2ax}$
• v0 is the initial velocity, v is the final velocity, a is the acceleration and x is the distance traveled
2. That means, an object was moving with an initial velocity of v0
• It was given an acceleration of a
♦ By Newtons's second law, acceleration is achieved by the application of a force F
• As a result, it's velocity changed to v
• During the 'time interval in which this change in velocity occurred', the object traveled a distance x
• This is shown in fig.6.6 below:
3. Let the mass of the object be m
• Multiplying both sides by m⁄2, we get:
$\mathbf\small{\frac{1}{2}mv^2-\frac{1}{2}mv_0^2=max}$
4. From Newton's second law, (m×a) = F
So we get: $\mathbf\small{\frac{1}{2}mv^2-\frac{1}{2}mv_0^2=Fx}$
5. We know that, to cause an acceleration, a force is required.
■ So it is obvious that, the 'F' in step (4) is the force which caused the acceleration 'a' which in turn changed the velocity from v0 to v
6. Let us analyze the equation in (4) in some detail:
• On the right side we have 'Fx'. It is Force × displacement
• But 'Force × displacement' is work done by the force
• So the right side of the equation in (4) gives the 'amount of work done by F'
7. Then the left side also must be 'work done'
■ How is that possible?
• It is simple actually. The work done by F is stored in the object as 'energy'
• The second term on the left side in (4) is $\mathbf\small{\frac{1}{2}mv_0^2}$
♦ This is the energy which the object posessed when it was moving with velocity v0.
• The first term on the left side is $\mathbf\small{\frac{1}{2}mv^2}$
♦ This is the energy which the object is possessing when it is moving moving with velocity v.
8. The difference $\mathbf\small{\frac{1}{2}mv^2-\frac{1}{2}mv_0^2}$ is the 'change in energy' acquired by the object
• The object was able to acquire this 'change in energy' because, work was done on it by F
• The 'change in energy' acquired is exactly equal to the 'work done by F'. Hence we have the '=' sign between the LHS and RHS in (4)
9. $\mathbf\small{\frac{1}{2}mv_0^2}$ is the initial energy
• $\mathbf\small{\frac{1}{2}mv^2}$ is the final energy
■ The energy posessed by an object due to it's motion is called kinetic energy
So we can write:
• $\mathbf\small{\frac{1}{2}mv_0^2}$ is the initial kinetic energy
• $\mathbf\small{\frac{1}{2}mv^2}$ is the final kinetic energy
■ If an object of mass 'm' is moving with a velocity 'v', then the kinetic energy pocessed by that object is $\mathbf\small{\frac{1}{2}mv^2}$
10. We see many examples of kinetic energy in our day to day life
• A 'pebble A' moving with a certain velocity possesses kinetic energy. It can hit another 'pebble B' and displace it.
• If 'pebble A' is stationary, it does not possess kinetic energy. It will not be able to displace 'pebble B'
It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass 1.00 g falling from a height of 1 km. It hits the ground with a speed of 50.0 m s-1. (a) What is the work done by the gravitational force? (b) What is the work done by the unknown resistive force?
Solution:
1. Rain drop is formed in the upper atmosphere when the surrounding temperature falls. The water vapor then condenses to form liquid water droplets. These water droplets combine together to form a larger water drop. Once a drop is formed, gravity will pull it towards the earth.
• So we see that the drop is in free fall. And it's initial velocity v0 is zero
2. The drop in our problem falls from a height of 1 km (1000 m)
• We know that velocity attained during free fall do not depend on mass. It depends only on the height of fall
• So how much velocity will be attained by the drop when it reaches the ground?
• We can find this in two steps:
(i) We have x = v0t + 1⁄2at2
• Entering known values, we get: 1000 = 0 + 1⁄2 × 10 × t2
• So we get t = √200 = 10√2 seconds
(ii) We have v = v0 + at
• Entering known values, we get: v = 0 + 10 × 10√2 = 100√2 ms-1
3. But it is given that, the drop hits the ground with a velocity of 50 ms-1.
• That means the drop was not able to attain 100√2 ms-1.
• This is because the air resistance acted in the upward direction thus opposing the motion of the drop
• So two forces acted on the drop
(i) The downward gravitational force
(ii) The upward resistive force
4. When a force act on an object and moves it through a certain distance, work is being done by the force on that object
• This work will be stored in the object as kinetic energy
• So the gravitational force will give the drop a kinetic energy equal to 1⁄2mv2
= 1⁄2 × 0.001 × (100√2)2 = 10 joules
5. But this much kinetic energy was not attained. The velocity attained was only 50 ms-1.
• So actual kinetic energy attained = 1⁄2 × 0.001 × (50)2 = 1.25 joules
6. The difference occurred because, the air resistance did some work in the opposite direction
• We can write:
Work done by gravity - Work done by air resistance = Actual kinetic energy attained
• Entering known values, we get: 10 - Work done by air resistance = 1.25
• So 'Work done by air resistance' = (10-1.25) = 8.75 joules
• Note: After discussing about 'potential energy', we will do this problem again using those concepts also.
• In one dimensional motion, we do not need to use vector notations
• Next we will consider two dimensional motion
1. Consider fig.6.7 below:
• A force $\mathbf\small{\vec{F}}$ is acting at an angle θ to the horizontal
1. In chapter 3, we saw the third equation of motion:
$\mathbf\small{v^2-v_0^2=2ax}$
• v0 is the initial velocity, v is the final velocity, a is the acceleration and x is the distance traveled
2. That means, an object was moving with an initial velocity of v0
• It was given an acceleration of a
♦ By Newtons's second law, acceleration is achieved by the application of a force F
• As a result, it's velocity changed to v
• During the 'time interval in which this change in velocity occurred', the object traveled a distance x
• This is shown in fig.6.6 below:
 |
| Fig.6.6 |
• Multiplying both sides by m⁄2, we get:
$\mathbf\small{\frac{1}{2}mv^2-\frac{1}{2}mv_0^2=max}$
4. From Newton's second law, (m×a) = F
So we get: $\mathbf\small{\frac{1}{2}mv^2-\frac{1}{2}mv_0^2=Fx}$
5. We know that, to cause an acceleration, a force is required.
■ So it is obvious that, the 'F' in step (4) is the force which caused the acceleration 'a' which in turn changed the velocity from v0 to v
6. Let us analyze the equation in (4) in some detail:
• On the right side we have 'Fx'. It is Force × displacement
• But 'Force × displacement' is work done by the force
• So the right side of the equation in (4) gives the 'amount of work done by F'
7. Then the left side also must be 'work done'
■ How is that possible?
• It is simple actually. The work done by F is stored in the object as 'energy'
• The second term on the left side in (4) is $\mathbf\small{\frac{1}{2}mv_0^2}$
♦ This is the energy which the object posessed when it was moving with velocity v0.
• The first term on the left side is $\mathbf\small{\frac{1}{2}mv^2}$
♦ This is the energy which the object is possessing when it is moving moving with velocity v.
8. The difference $\mathbf\small{\frac{1}{2}mv^2-\frac{1}{2}mv_0^2}$ is the 'change in energy' acquired by the object
• The object was able to acquire this 'change in energy' because, work was done on it by F
• The 'change in energy' acquired is exactly equal to the 'work done by F'. Hence we have the '=' sign between the LHS and RHS in (4)
9. $\mathbf\small{\frac{1}{2}mv_0^2}$ is the initial energy
• $\mathbf\small{\frac{1}{2}mv^2}$ is the final energy
■ The energy posessed by an object due to it's motion is called kinetic energy
So we can write:
• $\mathbf\small{\frac{1}{2}mv_0^2}$ is the initial kinetic energy
• $\mathbf\small{\frac{1}{2}mv^2}$ is the final kinetic energy
■ If an object of mass 'm' is moving with a velocity 'v', then the kinetic energy pocessed by that object is $\mathbf\small{\frac{1}{2}mv^2}$
10. We see many examples of kinetic energy in our day to day life
• A 'pebble A' moving with a certain velocity possesses kinetic energy. It can hit another 'pebble B' and displace it.
• If 'pebble A' is stationary, it does not possess kinetic energy. It will not be able to displace 'pebble B'
The work-energy (WE) theorem states that: The change in kinetic energy of a particle is equal to the work done on it by the net force
Solved example 6.4
Solution:
1. Rain drop is formed in the upper atmosphere when the surrounding temperature falls. The water vapor then condenses to form liquid water droplets. These water droplets combine together to form a larger water drop. Once a drop is formed, gravity will pull it towards the earth.
• So we see that the drop is in free fall. And it's initial velocity v0 is zero
2. The drop in our problem falls from a height of 1 km (1000 m)
• We know that velocity attained during free fall do not depend on mass. It depends only on the height of fall
• So how much velocity will be attained by the drop when it reaches the ground?
• We can find this in two steps:
(i) We have x = v0t + 1⁄2at2
• Entering known values, we get: 1000 = 0 + 1⁄2 × 10 × t2
• So we get t = √200 = 10√2 seconds
(ii) We have v = v0 + at
• Entering known values, we get: v = 0 + 10 × 10√2 = 100√2 ms-1
3. But it is given that, the drop hits the ground with a velocity of 50 ms-1.
• That means the drop was not able to attain 100√2 ms-1.
• This is because the air resistance acted in the upward direction thus opposing the motion of the drop
• So two forces acted on the drop
(i) The downward gravitational force
(ii) The upward resistive force
4. When a force act on an object and moves it through a certain distance, work is being done by the force on that object
• This work will be stored in the object as kinetic energy
• So the gravitational force will give the drop a kinetic energy equal to 1⁄2mv2
= 1⁄2 × 0.001 × (100√2)2 = 10 joules
5. But this much kinetic energy was not attained. The velocity attained was only 50 ms-1.
• So actual kinetic energy attained = 1⁄2 × 0.001 × (50)2 = 1.25 joules
6. The difference occurred because, the air resistance did some work in the opposite direction
• We can write:
Work done by gravity - Work done by air resistance = Actual kinetic energy attained
• Entering known values, we get: 10 - Work done by air resistance = 1.25
• So 'Work done by air resistance' = (10-1.25) = 8.75 joules
• Note: After discussing about 'potential energy', we will do this problem again using those concepts also.
• In the discussions so far in this section, the motions were all rectilinear (that is., one dimensional)
• Next we will consider two dimensional motion
1. Consider fig.6.7 below:
 |
| Fig.6.7 |
2. Because of this angle, the block will tend to move in two directions:
(i) horizontally to the right
(ii) vertically towards bottom
• In this case, there will not be a motion downwards. This is due to the presence of the floor.
• Even then, because of the slope of the force, it is a case of two dimensional motion and we have to use vector notations
3. Because of the slope of the force, there will be two force components
• One vertical and the other horizontal
• The horizontal component causes the block to move horizontally
• The vertical component is unable to cause any movement in the vertical direction because of the floor
4. So what is the work done?
• We see that the displacement is in the horizontal direction
• Word done is (Force × displacement)
• But to use this equation, the force must be in the same direction as the displacement
5. So we must take the horizontal component of the force
• Magnitude of the horizontal component is $\mathbf\small{|\vec{F}|\cos \theta}$
6. If the displacement vector is $\mathbf\small{\vec{d_x}}$, then the magnitude of displacement is $\mathbf\small{|\vec{d_x}|}$
7. So work done = $\mathbf\small{|\vec{F}|\cos \theta \times |\vec{d_x}|}$
• This can be rearranged and written as:
• Work done = $\mathbf\small{|\vec{F}||\vec{d_x}|\cos \theta }$
• Note that, θ is the angle between $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d_x}}$
■ So, in the right side of the above equation, what we have is the dot product of two vectors $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d_x}}$
■ Thus we can write:
Work done = $\mathbf\small{\vec{F}.\vec{d_x}}$
That is., work done is the dot product of force vector and displacement vector.
8. In our present case, since there is no displacement in the vertical direction, there is no work done by the vertical component of the force
• But if the floor was absent, there would be work done in vertical direction also
• Can we use dot product to find that work?
9. Consider fig.6.8 below:
• Since the floor is absent, the object moves both in horizontally and vertically
• We want the work done by the vertical component of $\mathbf\small{\vec{F}}$
• $\mathbf\small{\vec{F}}$ makes an angle θ1 with the vertical
• So magnitude of the vertical component of $\mathbf\small{\vec{F}}$ = $\mathbf\small{|\vec{F}|\cos \theta_1}$
10. If the displacement vector in the vertical direction is $\mathbf\small{\vec{d_y}}$, then the magnitude of displacement is $\mathbf\small{|\vec{d_y}|}$
11. So work done = $\mathbf\small{|\vec{F}|\cos \theta_1 \times |\vec{d_y}|}$
• This can be rearranged and written as:
• Work done = $\mathbf\small{|\vec{F}||\vec{d_y}|\cos \theta_1 }$
• Note that, θ1 is the angle between $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d_y}}$
■ So, in the right side of the above equation, what we have is the dot product of two vectors $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d_y}}$
■ Thus we can write:
In the vertical direction also, work done is the dot product
12. When ever we take the product of $\mathbf\small{|\vec{F}|}$ and cos θ, we are getting the component of that force along the direction of displacement
■ So we can write:
Work done by a force is the product of two items:
(i) Component of the force in the direction of displacement
(ii) Magnitude of the displacement
■ In equation form, we get:
Eq.6.15:. $\mathbf\small{W=\vec{F}.\vec{d}}$
From Eq.6.15, it is clear that, if displacement is zero, work done is zero
An example:
• If a person push hard against a rigid wall, there is no displacement for the wall.
♦ Then the work done by the person is zero
• The person will get tired after some time. This is due to the alternate contracting and relaxing of the muscles. And also due to the usage of internal energy.
• So the meaning of work in physics is different from it’s usage in everyday language.
Another example:
• A weight lifter holding a 150 kg mass steadily for 30 s does no work on that mass
♦ This is because there is no displacement for that mass
Another example:
• A block may move large distance on a smooth floor.
• This movement will not require any force since there is no friction to overcome.
• In such a motion, displacement is large.
• But force is zero.
♦ Then work done is also zero
An interesting example:
• We have seen that gravitational force does work on a rain drop.
• Now consider this: A block of mass m is moving horizontally on a smooth floor.
• The gravitational force mg is acting continuously on the block.
• So there is both force and displacement.
• But the force is vertical and displacement is horizontal.
♦ The angle between the force vector and displacement vector is 90o
• Then the dot product becomes zero because cos 90 = 0
• Thus in this case, we can say: the gravitational force does not do any work on the block
Another example:
• Assume that the moon’s orbit around the earth is a perfect circle
• Then the gravitational force between the earth and the moon will be in the radial direction
• The motion of the moon at any instant is tangential to the orbit
• We know that in any circle, the radius is perpendicular to tangent
• This is shown in fig.6.9(a) below:
• So the direction of gravitational force is perpendicular to the direction of displacement of the moon
• Thus we can say: the work done by gravitational force on the moon is zero
• But the force is acting in a sloping direction towards the left
2. In this case also, to find the work done, we take the same dot product: $\mathbf\small{\vec{F}.\vec{d}=|\vec{F}||\vec{d}|\cos \theta}$
• But here θ is greater than 90o
• From trigonometry classes we know that when θ is between 90o and 180o, cos θ becomes negative.
♦ For example, cos 150 = -0.8660
• So the final dot product becomes negative
■ That means the work done is negative
• This is understandable because force is acting in a direction opposite to that of the displacement.
1. We have: work = Force × displacement
2. Force = Mass × acceleration
3. Acceleration = change in velocity per second
4. velocity = distance/time
• So working on the above steps from velocity upwards, we get:
1. Velocity has dimensions: [LT-1]
2. Acceleration has dimensions: [LT-2]
3. Force has dimensions: [MLT-2]
4. Work has dimensions: [ML2T-2]
A cyclist comes to a skidding stop in 10 m. During this process the force on the cycle due to the road is 200 N and is directly opposed to the motion. (a) How much work does the road do on the cycle? (b) How much work does the cycle do on the road?
Solution:
• The cyclist stops pedaling and applies the brakes
• If he was pedaling at a normal speed, he could bring the cycle to a stop without skidding
• If the speed was large, and brakes are applied suddenly, wheels would stop spinning instantly and the cycle would skid. The cyclist in our problem encountered this situation
• Now we can write the steps:
Part (a):
1. When skidding occurs, road applies frictional force on the cycle
• It is given that the magnitude of this force $\mathbf\small{\vec{F}}$ is 200 N
2. Also it is given that the force acted for a distance of 10 m
• So magnitude of the displacement vector $\mathbf\small{\vec{d}}$ is 10 m
3. If we assume that the displacement vector to be from left to right, then force vector will be from right to left
• This is because, the frictional force is always opposite to the direction of motion
• Thus the angle θ between $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d}}$ is 180o
4. So the magnitude of the work done = $\mathbf\small{\vec{F}.\vec{d}=|\vec{F}||\vec{d}|\cos \theta}$
= 200 × 10 × cos 180
= 200 × 10 × -1
= -2000 J
5. It is this negative energy that brings the cycle to a stop
• Let us apply the work-energy theorem to this situation:
(i) horizontally to the right
(ii) vertically towards bottom
• In this case, there will not be a motion downwards. This is due to the presence of the floor.
• Even then, because of the slope of the force, it is a case of two dimensional motion and we have to use vector notations
3. Because of the slope of the force, there will be two force components
• One vertical and the other horizontal
• The horizontal component causes the block to move horizontally
• The vertical component is unable to cause any movement in the vertical direction because of the floor
4. So what is the work done?
• We see that the displacement is in the horizontal direction
• Word done is (Force × displacement)
• But to use this equation, the force must be in the same direction as the displacement
5. So we must take the horizontal component of the force
• Magnitude of the horizontal component is $\mathbf\small{|\vec{F}|\cos \theta}$
6. If the displacement vector is $\mathbf\small{\vec{d_x}}$, then the magnitude of displacement is $\mathbf\small{|\vec{d_x}|}$
7. So work done = $\mathbf\small{|\vec{F}|\cos \theta \times |\vec{d_x}|}$
• This can be rearranged and written as:
• Work done = $\mathbf\small{|\vec{F}||\vec{d_x}|\cos \theta }$
• Note that, θ is the angle between $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d_x}}$
■ So, in the right side of the above equation, what we have is the dot product of two vectors $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d_x}}$
■ Thus we can write:
Work done = $\mathbf\small{\vec{F}.\vec{d_x}}$
That is., work done is the dot product of force vector and displacement vector.
8. In our present case, since there is no displacement in the vertical direction, there is no work done by the vertical component of the force
• But if the floor was absent, there would be work done in vertical direction also
• Can we use dot product to find that work?
9. Consider fig.6.8 below:
 |
| Fig.6.8 |
• We want the work done by the vertical component of $\mathbf\small{\vec{F}}$
• $\mathbf\small{\vec{F}}$ makes an angle θ1 with the vertical
• So magnitude of the vertical component of $\mathbf\small{\vec{F}}$ = $\mathbf\small{|\vec{F}|\cos \theta_1}$
10. If the displacement vector in the vertical direction is $\mathbf\small{\vec{d_y}}$, then the magnitude of displacement is $\mathbf\small{|\vec{d_y}|}$
11. So work done = $\mathbf\small{|\vec{F}|\cos \theta_1 \times |\vec{d_y}|}$
• This can be rearranged and written as:
• Work done = $\mathbf\small{|\vec{F}||\vec{d_y}|\cos \theta_1 }$
• Note that, θ1 is the angle between $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d_y}}$
■ So, in the right side of the above equation, what we have is the dot product of two vectors $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d_y}}$
■ Thus we can write:
In the vertical direction also, work done is the dot product
12. When ever we take the product of $\mathbf\small{|\vec{F}|}$ and cos θ, we are getting the component of that force along the direction of displacement
■ So we can write:
Work done by a force is the product of two items:
(i) Component of the force in the direction of displacement
(ii) Magnitude of the displacement
■ In equation form, we get:
Eq.6.15:. $\mathbf\small{W=\vec{F}.\vec{d}}$
From Eq.6.15, it is clear that, if displacement is zero, work done is zero
An example:
• If a person push hard against a rigid wall, there is no displacement for the wall.
♦ Then the work done by the person is zero
• The person will get tired after some time. This is due to the alternate contracting and relaxing of the muscles. And also due to the usage of internal energy.
• So the meaning of work in physics is different from it’s usage in everyday language.
Another example:
• A weight lifter holding a 150 kg mass steadily for 30 s does no work on that mass
♦ This is because there is no displacement for that mass
Another example:
• A block may move large distance on a smooth floor.
• This movement will not require any force since there is no friction to overcome.
• In such a motion, displacement is large.
• But force is zero.
♦ Then work done is also zero
An interesting example:
• We have seen that gravitational force does work on a rain drop.
• Now consider this: A block of mass m is moving horizontally on a smooth floor.
• The gravitational force mg is acting continuously on the block.
• So there is both force and displacement.
• But the force is vertical and displacement is horizontal.
♦ The angle between the force vector and displacement vector is 90o
• Then the dot product becomes zero because cos 90 = 0
• Thus in this case, we can say: the gravitational force does not do any work on the block
Another example:
• Assume that the moon’s orbit around the earth is a perfect circle
• Then the gravitational force between the earth and the moon will be in the radial direction
• The motion of the moon at any instant is tangential to the orbit
• We know that in any circle, the radius is perpendicular to tangent
• This is shown in fig.6.9(a) below:
 |
| Fig.6.9 |
• Thus we can say: the work done by gravitational force on the moon is zero
Negative work
1. Consider the block in fig.6.9(b) above. The displacement of the block is horizontal and towards the right.• But the force is acting in a sloping direction towards the left
2. In this case also, to find the work done, we take the same dot product: $\mathbf\small{\vec{F}.\vec{d}=|\vec{F}||\vec{d}|\cos \theta}$
• But here θ is greater than 90o
• From trigonometry classes we know that when θ is between 90o and 180o, cos θ becomes negative.
♦ For example, cos 150 = -0.8660
• So the final dot product becomes negative
■ That means the work done is negative
• This is understandable because force is acting in a direction opposite to that of the displacement.
Fig.6.9(b) above can be used to discuss a few more details about the 'influence of θ on work done'
We will write the steps:
1. In the fig.6.9(b), when θ = 0, $\mathbf\small{\vec{F}}$ will be perfectly horizontal.
• And also it will be acting towards the right
• When θ = 0, cos θ = 1
• Thus '$\mathbf\small{|\vec{F}||\vec{d}|}$' will be multiplied by '1'
• So the final dot product $\mathbf\small{|\vec{F}||\vec{d}|\cos \theta}$ will be having the maximum possible value
■ That means when θ = 0, the 'work done' will be the maximum
■ In general, when θ = 0, the dot product of two vectors will be having the maximum possible value
2. In the fig.6.9(b), when θ is greater than zero but less than 90, $\mathbf\small{\vec{F}}$ will be sloping towards the right
• When θ is greater than zero but less than 90, cos θ will be a 'positive fraction'
♦ For example, cos 40 = 0.7660
• Thus '$\mathbf\small{|\vec{F}||\vec{d}|}$' will be multiplied by a 'positive fraction'
• So the final dot product $\mathbf\small{|\vec{F}||\vec{d}|\cos \theta}$ will be positive but less than maximum
3. In the fig.6.9(b), when θ = 180, $\mathbf\small{\vec{F}}$ will be perfectly horizontal.
• And also it will be acting towards the left
• When θ = 180, cos θ = -1
• Thus '$\mathbf\small{|\vec{F}||\vec{d}|}$' will be multiplied by '-1'
• So the final dot product $\mathbf\small{|\vec{F}||\vec{d}|\cos \theta}$ will be having the 'maximum possible magnitude', but in the opposite direction
■ The work done is opposite to the direction of motion. Such work is considered as 'negative work'
• The work done by friction is an example of this case. On a horizontal surface, the angle between the displacement vector and the 'frictional force vector' will be 180o.
■ In general, when θ = 180, the dot product of two vectors will be having the least possible value because of the 'negative sign'
4. In the fig.6.9(b), when θ is greater than 90 but less than 180, $\mathbf\small{\vec{F}}$ will be sloping towards the left
• When θ is greater than 90 but less than 180, cos θ will be a 'negative fraction'
♦ For example, cos 135 = -0.7071
• Thus '$\mathbf\small{|\vec{F}||\vec{d}|}$' will be multiplied by a 'negative fraction'
• So the final dot product $\mathbf\small{|\vec{F}||\vec{d}|\cos \theta}$ will be negative but greater than the least work in (3)
We will write the steps:
1. In the fig.6.9(b), when θ = 0, $\mathbf\small{\vec{F}}$ will be perfectly horizontal.
• And also it will be acting towards the right
• When θ = 0, cos θ = 1
• Thus '$\mathbf\small{|\vec{F}||\vec{d}|}$' will be multiplied by '1'
• So the final dot product $\mathbf\small{|\vec{F}||\vec{d}|\cos \theta}$ will be having the maximum possible value
■ That means when θ = 0, the 'work done' will be the maximum
■ In general, when θ = 0, the dot product of two vectors will be having the maximum possible value
2. In the fig.6.9(b), when θ is greater than zero but less than 90, $\mathbf\small{\vec{F}}$ will be sloping towards the right
• When θ is greater than zero but less than 90, cos θ will be a 'positive fraction'
♦ For example, cos 40 = 0.7660
• Thus '$\mathbf\small{|\vec{F}||\vec{d}|}$' will be multiplied by a 'positive fraction'
• So the final dot product $\mathbf\small{|\vec{F}||\vec{d}|\cos \theta}$ will be positive but less than maximum
3. In the fig.6.9(b), when θ = 180, $\mathbf\small{\vec{F}}$ will be perfectly horizontal.
• And also it will be acting towards the left
• When θ = 180, cos θ = -1
• Thus '$\mathbf\small{|\vec{F}||\vec{d}|}$' will be multiplied by '-1'
• So the final dot product $\mathbf\small{|\vec{F}||\vec{d}|\cos \theta}$ will be having the 'maximum possible magnitude', but in the opposite direction
■ The work done is opposite to the direction of motion. Such work is considered as 'negative work'
• The work done by friction is an example of this case. On a horizontal surface, the angle between the displacement vector and the 'frictional force vector' will be 180o.
■ In general, when θ = 180, the dot product of two vectors will be having the least possible value because of the 'negative sign'
4. In the fig.6.9(b), when θ is greater than 90 but less than 180, $\mathbf\small{\vec{F}}$ will be sloping towards the left
• When θ is greater than 90 but less than 180, cos θ will be a 'negative fraction'
♦ For example, cos 135 = -0.7071
• Thus '$\mathbf\small{|\vec{F}||\vec{d}|}$' will be multiplied by a 'negative fraction'
• So the final dot product $\mathbf\small{|\vec{F}||\vec{d}|\cos \theta}$ will be negative but greater than the least work in (3)
Dimensions of work
2. Force = Mass × acceleration
3. Acceleration = change in velocity per second
4. velocity = distance/time
• So working on the above steps from velocity upwards, we get:
1. Velocity has dimensions: [LT-1]
2. Acceleration has dimensions: [LT-2]
3. Force has dimensions: [MLT-2]
4. Work has dimensions: [ML2T-2]
Solved example 6.5
Solution:
• The cyclist stops pedaling and applies the brakes
• If he was pedaling at a normal speed, he could bring the cycle to a stop without skidding
• If the speed was large, and brakes are applied suddenly, wheels would stop spinning instantly and the cycle would skid. The cyclist in our problem encountered this situation
• Now we can write the steps:
Part (a):
1. When skidding occurs, road applies frictional force on the cycle
• It is given that the magnitude of this force $\mathbf\small{\vec{F}}$ is 200 N
2. Also it is given that the force acted for a distance of 10 m
• So magnitude of the displacement vector $\mathbf\small{\vec{d}}$ is 10 m
3. If we assume that the displacement vector to be from left to right, then force vector will be from right to left
• This is because, the frictional force is always opposite to the direction of motion
• Thus the angle θ between $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d}}$ is 180o
4. So the magnitude of the work done = $\mathbf\small{\vec{F}.\vec{d}=|\vec{F}||\vec{d}|\cos \theta}$
= 200 × 10 × cos 180
= 200 × 10 × -1
= -2000 J
5. It is this negative energy that brings the cycle to a stop
• Let us apply the work-energy theorem to this situation:
■ The work-energy (WE) theorem states that: The change in kinetic energy of a particle is equal to the work done on it by the net force
• The change in kinetic energy = k2 -k1
• The final kinetic energy k2 is zero
• So the change in kinetic energy is negative
• That means the work done by the net force (frictional force) is negative
Part (b):
1. The road applies a force of 200 N on the cycle
2. By Newton's third law, the cycle will apply a force of 200 N on the road
3. But the road does not undergo any displacement
• So work done by the cycle on the road is zero
4. We can write:
• According to Newton's third law, the following two forces will be always equal and opposite:
(i) Force applied by a body A on another body B
(ii) The reaction force applied by B on A
• But the following works need not be equal and opposite:
(i) Work done by a body A on another body B
(ii) Work done by the 'reaction force of B' on A
Solved examples 6.6
A force $\mathbf\small{\vec{F}=2\hat{i}+3\hat{j}+4\hat{k}}$ N causes displacement of an object. The position vector of the initial position of the object is $\mathbf\small{\vec{r_1}=2\hat{i}+3\hat{j}+1\hat{k}}$ m. The position vector of the final position is $\mathbf\small{\vec{r_2}=\hat{i}+\hat{j}+\hat{k}}$ m. What is the work done by $\mathbf\small{\vec{F}}$?
Solution:
1. Displacement vector $\mathbf\small{\vec{d}=\vec{r_2}-\vec{r_1}}$ (Details here)
• Thus we get: $\mathbf\small{\vec{d}=(\hat{i}+\hat{j}+\hat{k})-(2\hat{i}+3\hat{j}+\hat{k})=(-\hat{i}-2\hat{j})}$ m
2. Work done = $\mathbf\small{\vec{F}.\vec{d}=F_xd_x+F_yd_y+F_zd_z}$ (Using Eq.6.8)
= (2 × -1 + 3 × -2) = (-2-6) = -8 J
• The change in kinetic energy = k2 -k1
• The final kinetic energy k2 is zero
• So the change in kinetic energy is negative
• That means the work done by the net force (frictional force) is negative
Part (b):
1. The road applies a force of 200 N on the cycle
2. By Newton's third law, the cycle will apply a force of 200 N on the road
3. But the road does not undergo any displacement
• So work done by the cycle on the road is zero
4. We can write:
• According to Newton's third law, the following two forces will be always equal and opposite:
(i) Force applied by a body A on another body B
(ii) The reaction force applied by B on A
• But the following works need not be equal and opposite:
(i) Work done by a body A on another body B
(ii) Work done by the 'reaction force of B' on A
Solved examples 6.6
A force $\mathbf\small{\vec{F}=2\hat{i}+3\hat{j}+4\hat{k}}$ N causes displacement of an object. The position vector of the initial position of the object is $\mathbf\small{\vec{r_1}=2\hat{i}+3\hat{j}+1\hat{k}}$ m. The position vector of the final position is $\mathbf\small{\vec{r_2}=\hat{i}+\hat{j}+\hat{k}}$ m. What is the work done by $\mathbf\small{\vec{F}}$?
Solution:
1. Displacement vector $\mathbf\small{\vec{d}=\vec{r_2}-\vec{r_1}}$ (Details here)
• Thus we get: $\mathbf\small{\vec{d}=(\hat{i}+\hat{j}+\hat{k})-(2\hat{i}+3\hat{j}+\hat{k})=(-\hat{i}-2\hat{j})}$ m
2. Work done = $\mathbf\small{\vec{F}.\vec{d}=F_xd_x+F_yd_y+F_zd_z}$ (Using Eq.6.8)
= (2 × -1 + 3 × -2) = (-2-6) = -8 J
No comments:
Post a Comment