In the previous section, we saw the basics about work done. In this section, we will see details about kinetic energy.
• The kinetic energy of an object (of mass m) moving with velocity is given by $\mathbf\small{K=\frac{1}{2}m \times\vec{v}.\vec{v}}$
♦ Using Eq.6.2 , we get: $\mathbf\small{\vec{v}.\vec{v}=|\vec{v}|^2}$
• Thus we get Eq.6.16: $\mathbf\small{K=\frac{1}{2}m |\vec{v}|^2}$
• $\mathbf\small{|\vec{v}|}$ is a 'magnitude'. That is., it is a simple scalar quantity (an orninary number). So $\mathbf\small{|\vec{v}|^2}$ can be easily calculated.
■ Kinetic energy is a scalar quantity. It is the ‘amount of work’ that an object can do by virtue of it’s motion
• For example, the kinetic energy of running water has been used from ancient times to grind corn
• The kinetic energy of winds can be used for sailing ships
• A bullet though small, can pierce objects because it has large kinetic energy
Solved example 6.7
In a ballistics demonstration a police officer fires a bullet of mass 50.0 g with speed 200 ms-1 on soft plywood of thickness 2.00 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet ?
Solution:
1. Initial kinetic energy = $\mathbf\small{K_1=\frac{1}{2}m |\vec{v_1}|^2=\frac{1}{2} \times \frac{50}{1000} \times 200^2= 1000\;J}$
2. Final kinetic energy = 10% of K1 = 1000 × 0.1 = 100 J
• But final kinetic energy = $\mathbf\small{K_2=\frac{1}{2}m |\vec{v_2}|^2}$
$\mathbf\small{=\frac{1}{2} \times \frac{50}{1000} \times |\vec{v_2}|^2 = 0.025 \times |\vec{v_2}|^2}$
3. So we can write: $\mathbf\small{100=0.025 \times |\vec{v_2}|^2}$
Thus we get: $\mathbf\small{|\vec{v_2}|=63.24\;ms^{-1}}$
4. We have: $\mathbf\small{\frac{|\vec{v_2}|}{|\vec{v_1}|}\times 100 =\frac{63.24}{200}\times 100 =31.62\,\text{%}}$
• That means v2 is about 32% of v1
5. Given that K2 is 10% of K1.
• So we would expect v2 to be 10% of v1.
• But we find that such a drastic reduction in velocity does not occur
• Velocity is reduced to 32% only. Not 10%
Solved example 6.8
An object of mass 30 kg is moving horizontally along a straight line, with a velocity of 0.5 ms-1. When it reaches point A, a force of 120 N pushes it through a distance of AB equal to 0.8 m. How much velocity does the object acquire when it reaches B? How far does it travel after B?
Take frictional force to be 5 N
Solution:
Part (a):
1. When the object is at A, it's velocity is 0.5 ms-1
So $\mathbf\small{K_A=\frac{1}{2}m|\vec{v_A}|^2=0.5 \times 30 \times 0.5^2 =3.75\;J}$
2. Work done = Net Force × distance = (120-5) × 0.8 = 92 J
• From the work-energy theorem, we have: KB - KA = Work done
3. Entering known values, we get: KB - 3.75 = 92 J
• Thus we get: KB = 92+3.75 = 95.75 J
4. We have: $\mathbf\small{K_B=\frac{1}{2}m|\vec{v_B}|^2=0.5 \times 30 \times |\vec{v_B}|^2=95.75\;J}$
• From this we get: $\mathbf\small{|\vec{v_B}|=2.526 \; \text{ms}^{-1}}$
Part (b):
1. When the object reaches B, the pushing is stopped. So after B, the object travels on it's own
• The energy for this motion is the kinetic energy acquired at B
• This kinetic energy will be used up by doing work against friction
2. Let the distance moved after 'B' be 'x' m
• Then work done by friction = Frictional force × distance = 5x
3. This much energy will be extracted from the object
• Thus we can write: 95.75 = 5x
• So x = 19.15 m
Another method:
1. Consider the motion beyond 'B'
• The initial velocity v0 for this motion is 2.526
• The final velocity v for this motion is 0 ms-1
2. During this motion, the only force acting on the body is the frictional force which is equal to -5 N
• So acceleration = Force⁄mass = -5⁄30 = -1⁄6 ms-2
3. We can use the equation: $\mathbf\small{v^2-v_0^2=2ax}$
• Entering known values, we get: $\mathbf\small{0^2-2.526^2=2 \times \frac{-1}{6} \times x}$
• From this we get: x = 19.142 m
• This result is same as the one we obtained earlier
Now we will discuss about the effect of varying force
First we will see a constant force. We will write the steps:
1. Consider the graph shown in fig.6.10(a) below
• Displacement x is plotted along the x axis
• Force F is plotted along the y axis
• So it is a Force-Displacement graph
2. Consider the cyan horizontal line
• It intersects the y axis at P. The y coordinate of P is F1
• So the cyan horizontal line indicates a constant force F1 acting on a body
3. Under the action of the constant force F1, the body is being continuously displaced
• Assume that the path along which the body is displaced is the x axis
4. On the x axis (that is., on the path of the body), two points A and B are marked
• The x coordinate of A is x1
• The x coordinate of B is x2
• Vertical red lines are drawn at A and B
5. Consider the ‘time interval in which the body traveled from A to B’
• We want the kinetic energy acquired by the body during this time interval
• That is., we want the work done on the body during the travel from A to B
6. We know that, work done is (force × displacement)
• So in our present case the work done from A to B is (F1 × ‘distance AB’)
7. Now consider fig.6.10(b) above
• It is the same graph in fig.a. The only difference is that, a particular area is shaded
8. We can see that the shaded area is enclosed between two sets of lines:
• First set of lines: Force graph and the x axis on top and bottom
• Second set of lines: Vertical red lines on the left and right
9. This shaded area is a rectangle
• The length of the rectangle is the ‘distance AB’
• Height of the rectangle is F1
• So area of the rectangle is (F1 × ‘distance AB’)
10. Comparing (6) and (9), we obtain the following relation:
■ Work done from A to B = Area enclosed between A and B
Now we will see a varying force.
1. In real life situations, the forces applied are not constant.
• That means, we do not get a perfect horizontal cyan line as in fig.6.10(a)
• The graph of the force will be a curve as shown in fig.6.11(a) below:
2. In the fig.6.11(a), we see that the force initially increases. Then it briefly becomes some what constant. After that it decreases.
• But overall, it is a curve, indicating that it is a varying force
3. The area enclosed by the curve is shown in fig.6.11(b)
• We want to prove that, this area shown in fig.6.11(b) is equal to the work done (from A to B) by the varying force.
4. Before proving that, we must device a method to find the area of such irregular figures. Let us try:
• The same area in fig.6.11(b) is shown in fig.6.12(a) below:
• But the area is divided into 6 strips
5. Those strips have two important features:
(i) Each strip is a perfect rectangle
(ii) The widths of all the strips are the same
6. Since any one strip is a perfect rectangle, we can calculate it’s area easily. Let us see how:
• If we divide the ‘length AB’ by 6, we will get the width of strip
• All strips have the same width. So we can call it the ‘common width’
• This ‘common width’ represents the ‘change in position x’
• So we can indicate the ‘common width’ by 'Δx' (Delta x)
■ We can say: All the strips have the same width Δx
7. Now to find the height of any one strip, look at it’s top-left corner
• That corner touches the curve
• So force ‘F’ at the ‘touching point’ of a rectangle can be taken as it’s height
8. The ‘touching point’ is different for different rectangles.
• Each rectangle has it’s own ‘unique value of F at the touching point’
• We can write:
♦ The value of F at the touching point of first rectangle is F1
♦ The value of F at the touching point of second rectangle is F2
♦ The value of F at the touching point of third rectangle is F3
♦ - - -
♦ - - -
♦ The value of F at the touching point of sixth rectangle is F6
9. Once we know the height of a rectangle, we can multiply it by the ‘common width Δx’ to calculate it’s area
• Thus we can calculate the 6 areas: A1, A2, A3, . . . , A6
♦ A1 = F1 × Δx
♦ A2 = F2 × Δx
♦ A3 = F3 × Δx
♦ - - -
♦ - - -
♦ A6 = F6 × Δx
10. Once we calculate all the 6 areas, we add them
• That is., we find the sum: A1 + A2 + A3 + . . . + A6
11. If all the areas were the same, we could have just written: Sum = 6A
• But here the areas are different
• We use a mathematical notation to express this type of summation
■ This notation is called ‘sigma notation’.
• It’s symbol is the Greek capital letter ($\mathbf\small{\Sigma}$)
• We write 'Sum of the 6 areas' as: $\mathbf\small{\sum_{i=1}^{i=6} A_i}$
11. Let us analyze the above notation:
• The ‘$\mathbf\small{A_i}$’ on the right side of ‘=’ sign conveys this information:
Certain ‘areas’ are to be added together
• How many areas are to be added?
• The answer is given by the ‘i=1’ and ‘i=6’ on bottom and top of $\mathbf\small{\Sigma}$
♦ ‘i= 1’ indicates A1
♦ ‘i= 6’ indicates A6
12. Thus we can write: $\mathbf\small{\sum_{i=1}^{i=6} A_i=A_1+A_2+\;.\;.\;.\;+A_6}$
• In the above expression, the right side is already understood and hence not written
■ So we can write:
The total area of the 6 rectangles in fig.6.12(a) is $\mathbf\small{\sum_{i=1}^{i=6} A_i}$
• We can expand $\mathbf\small{A_i}$.
♦ Expanding is possible because $\mathbf\small{A_i=(F_i \times \Delta x)}$
• We get:
The total area of the 6 rectangles in fig.6.12(a) = $\mathbf\small{\sum_{i=1}^{i=6} (F_i \times \Delta x)}$
13. So we successfully calculated the total area of the six rectangles
• But is this ‘total area’ equal to the shaded area in fig.6.11(b)?
■ No. It isn’t.
14. Let us see the reason:
• Looking closely at fig.6.12(a) we see these two facts:
(i) At the rising part of the cyan curve on the left, there are ‘black triangles’ just below the curve
(ii) At the falling part of the cyan curve on the right, there are ‘colored triangles’ just above the curve
• We have to add the areas of those 'black triangles' to the area in (12)
• Also we have to subtract the areas of those 'colored triangles' from the area in (12)
• Then only we will get a ‘perfect fit’ and thus a correct ‘total area’
16. Another important point can also be noted at this stage:
• Consider any one of the black triangles. One such triangle is shown enlarged in fig.6.12(b)
• It is named as triangle UVW
♦ U is the top-left corner of the rectangle. It is the 'touching point'
♦ V is the top-right corner of the rectangle
♦ W is the top-left corner of the adjacent rectangle
• We used the 'value of F' at the 'touching point' U as the height of that particular rectangle
• Note that, at the left point 'U' of the rectangle, the force acting is the 'force at U'
♦ We will indicate this as Fu
• But at the right point 'V', the force acting is the 'force at W'
♦ We will indicate this as Fw
• That means, during the displacement (Δx) from the bottom-left corner to bottom-right side of this rectangle, the force varies from Fu to Fw
■ That means, during the displacement (Δx) of this rectangle, the force is not constant.
• We will have to apply this information later in this discussion
17. Now to continue with our main discussion:
■ We need to improve our method so that:
Total area of the strips (rectangles) = Shaded area in fig.6.11(b)
• Let us try:
In fig.6.13(a) below, the same cyan curve is shown. The length AB is also the same
• But AB is now divided into 9 strips
• So the common width Δx will be less than the 'Δx in fig.6.12(a)'
18. The total area of the 9 rectangles is $\mathbf\small{\sum_{i=1}^{i=9} (F_i \times \Delta x)}$
• Here also we see black triangles and colored triangles
• So we can write:
Total area of the 9 strips (rectangles) ≠ Shaded area in fig.6.11(b)
19. But there is an improvement:
• The triangles have become smaller
• That means accuracy have increased. But not to the expected level
20. Let us try to improve further:
In fig.6.13(b) above, the same cyan curve is shown. The length AB is also the same
• But AB is now divided into 14 strips
• So the common width Δx will be less than the 'Δx in fig.6.13(a)'
21. The total area of the 14 rectangles is $\mathbf\small{\sum_{i=1}^{i=14} (F_i \times \Delta x)}$
• Here also we see black triangles and colored triangles
• So we can write:
Total area of the 14 strips (rectangles) ≠ Shaded area in fig.6.11(b)
22. But there is an improvement:
• The triangles have become still smaller
• That means accuracy have increased. But not to the expected level
23. Let us try to improve further:
In fig.6.14(a) below, the same cyan curve is shown. The length AB is also the same
• But AB is now divided into 19 strips
• So the common width Δx will be less than the 'Δx in fig.6.13(b)'
21. The total area of the 19 rectangles is $\mathbf\small{\sum_{i=1}^{i=19} (F_i \times \Delta x)}$
• Here also we see black triangles and colored triangles
• So we can write:
Total area of the 19 strips (rectangles) ≠ Shaded area in fig.6.11(b)
22. But there is much improvement:
• The triangles have become very small
• That means accuracy have increased very much. But not to the expected level
• We see that, with the increase in number of strips, the 'total area of strips' gets closer and closer to the 'required shaded area in fig.6.11(b)'
24. What if we make the 'number of strips' equal to infinity ($\mathbf\small{\infty}$)?
• Then the width of each strip will be $\mathbf\small{\frac{AB}{\infty}}$
• But any thing divided by $\mathbf\small{\infty}$ is zero.
• So $\mathbf\small{\frac{AB}{\infty}=0}$
25. That means, if we increase the 'number of strips' to $\mathbf\small{\infty}$, the widths of the strips will become zero
• Strips of 'zero width' are not helpful for calculating total area
26. So we need to prevent the widths from becoming zero
• And at the same time we need a very large number of strips also
• This situation is represented as: Total area of strips = $\mathbf\small{\lim_{i\to \infty}\sum A_i}$
27. Let us analyse the above expression:
• $\mathbf\small{\sum A_i}$ indicates that 'sum of areas' is to be calculated
• How many areas are there?
The answer is given by $\mathbf\small{\lim_{i\to \infty}}$
• 'i' tends to become infinity.
♦ If it becomes infinity, there is no use because strips will then have zero widths
• So 'i' is kept within a limit (indicated by 'lim')
• It indicates that 'i' should make a 'maximum allowable closeness' to infinity
28. In such a situation, the width of strips will have 'least possible values' like 0.00001 mm
• There can be any number of zeros between the 'decimal point' and '1'
• There is no restriction on that 'number of zeroes'
29. Now we can expand the expression in (26). We get:
• Total area of strips = $\mathbf\small{\lim_{\Delta x \,\to \,0}\sum_{x=x_i}^{x=x_f} (F_x \times \Delta x)}$
30. Let us analyze the above expression:
• $\mathbf\small{\sum (F_x \times \Delta x)}$ indicates that the following sum is to be calculated:
$\mathbf\small{(F_1 \times \Delta x)+(F_2 \times \Delta x)+(F_3 \times \Delta x)+\;.\;.\;.}$
• How many terms are there?
'That many terms' by which, the width Δx of each strip has 'a value very close to zero'
♦ This is indicated by: $\mathbf\small{\lim_{\Delta x \,\to \,0}}$
• The height of each strip is given by the force at the 'touching point'. It is denoted as $\mathbf\small{(F_x}$
♦ $\mathbf\small{F_{xi}}$ is the height of the first strip
♦ $\mathbf\small{F_{xf}}$ is the height of the last strip
31. We saw that, when the number of strips is increased, the triangles will become smaller and smaller
• Finally, when the number of strips approach infinity, those triangles will completely vanish
♦ This is shown in fig.6.14(b)
• That means, the area obtained in (29) is the required shaded area in fig.6.11(b)
32. Now we take up the point that was mentioned in (16)
• When the width of strips approach zero, the triangles vanish
• As shown in fig.6.14(b), the points U, V and W nearly coincide
• Then Fu is same as Fw.
• That means, the same force is acting at the left side and right side of the rectangle
33. This is applicable to all rectangles. We can write the following steps:
(i) Take any one rectangle
• Since Δx is so small, the force which displace the object from the bottom-left corner to the bottom-right corner of that rectangle is a constant
(ii) Since the force is constant for that rectangle, the work done on the object while displacing from the bottom-left corner to the bottom-right corner of that rectangle is given by: (Force corresponding to that rectangle × Δx)
(iii) But (Force corresponding to that rectangle × Δx) = Area of that rectangle
(iv) So 'sum of areas of all rectangles' is the total work done from A to B
34. 'Sum of areas of all rectangles' is given by (29). Thus we can write:
Sum of areas of all rectangles
= Area of the shaded portion in fig.6.11(b)
= Work done from A to B
= $\mathbf\small{\lim_{\Delta x \,\to \,0}\sum_{x=x_i}^{x=x_f} (F_x \times \Delta x)}$
35. So we need to add a very large number of rectangular areas
• But there is nothing to worry about. Such a summation can be done very easily using principles of calculus
• So we will revisit this topic after completing the basic math course on calculus
36. The body is moving along the x axis
• At the point 'A', it has a certain amount of kinetic energy
♦ It is given by: $\mathbf\small{K_A=\frac{1}{2}m|\vec{v_A}|^2}$
• Since the body is acted upon by a force, it will experience acceleration
• So it's velocity at B will be different. Consequently, the kinetic energy at B will also be different
♦ We can write: $\mathbf\small{K_B=\frac{1}{2}m|\vec{v_B}|^2}$
• According to the work-energy theorem, the change in kinetic energy is equal to the work done on the body from A to B
♦ But this work is equal to the area enclosed by the curve.
■ So we can write:
According to the work-energy theorem, the change in kinetic energy is equal to the area enclosed by the curve for a varying force
37. We will see the proof after completing the basic math course on calculus
At this stage, all we need to know are the following two points:
(i) Area enclosed by the 'curve representing a force' is equal to the work done by that force
• For a constant force, we proved this using fig.6.10 above
• For a varying force, we proved this using figs.6.11 to 6.14 above
(ii) 'Change in kinetic energy' acquired by the body is equal to the 'work done by the force' on that body
• For a constant force, we proved this in the previous section, using the equation $\mathbf\small{v^2-v_0^2=2ax}$
• For a varying force, we will see the proof after learning calculus
• The kinetic energy of an object (of mass m) moving with velocity is given by $\mathbf\small{K=\frac{1}{2}m \times\vec{v}.\vec{v}}$
♦ Using Eq.6.2 , we get: $\mathbf\small{\vec{v}.\vec{v}=|\vec{v}|^2}$
• Thus we get Eq.6.16: $\mathbf\small{K=\frac{1}{2}m |\vec{v}|^2}$
• $\mathbf\small{|\vec{v}|}$ is a 'magnitude'. That is., it is a simple scalar quantity (an orninary number). So $\mathbf\small{|\vec{v}|^2}$ can be easily calculated.
■ Kinetic energy is a scalar quantity. It is the ‘amount of work’ that an object can do by virtue of it’s motion
• For example, the kinetic energy of running water has been used from ancient times to grind corn
• The kinetic energy of winds can be used for sailing ships
• A bullet though small, can pierce objects because it has large kinetic energy
Solved example 6.7
In a ballistics demonstration a police officer fires a bullet of mass 50.0 g with speed 200 ms-1 on soft plywood of thickness 2.00 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet ?
Solution:
1. Initial kinetic energy = $\mathbf\small{K_1=\frac{1}{2}m |\vec{v_1}|^2=\frac{1}{2} \times \frac{50}{1000} \times 200^2= 1000\;J}$
2. Final kinetic energy = 10% of K1 = 1000 × 0.1 = 100 J
• But final kinetic energy = $\mathbf\small{K_2=\frac{1}{2}m |\vec{v_2}|^2}$
$\mathbf\small{=\frac{1}{2} \times \frac{50}{1000} \times |\vec{v_2}|^2 = 0.025 \times |\vec{v_2}|^2}$
3. So we can write: $\mathbf\small{100=0.025 \times |\vec{v_2}|^2}$
Thus we get: $\mathbf\small{|\vec{v_2}|=63.24\;ms^{-1}}$
4. We have: $\mathbf\small{\frac{|\vec{v_2}|}{|\vec{v_1}|}\times 100 =\frac{63.24}{200}\times 100 =31.62\,\text{%}}$
• That means v2 is about 32% of v1
5. Given that K2 is 10% of K1.
• So we would expect v2 to be 10% of v1.
• But we find that such a drastic reduction in velocity does not occur
• Velocity is reduced to 32% only. Not 10%
Solved example 6.8
An object of mass 30 kg is moving horizontally along a straight line, with a velocity of 0.5 ms-1. When it reaches point A, a force of 120 N pushes it through a distance of AB equal to 0.8 m. How much velocity does the object acquire when it reaches B? How far does it travel after B?
Take frictional force to be 5 N
Solution:
Part (a):
1. When the object is at A, it's velocity is 0.5 ms-1
So $\mathbf\small{K_A=\frac{1}{2}m|\vec{v_A}|^2=0.5 \times 30 \times 0.5^2 =3.75\;J}$
2. Work done = Net Force × distance = (120-5) × 0.8 = 92 J
• From the work-energy theorem, we have: KB - KA = Work done
3. Entering known values, we get: KB - 3.75 = 92 J
• Thus we get: KB = 92+3.75 = 95.75 J
4. We have: $\mathbf\small{K_B=\frac{1}{2}m|\vec{v_B}|^2=0.5 \times 30 \times |\vec{v_B}|^2=95.75\;J}$
• From this we get: $\mathbf\small{|\vec{v_B}|=2.526 \; \text{ms}^{-1}}$
Part (b):
1. When the object reaches B, the pushing is stopped. So after B, the object travels on it's own
• The energy for this motion is the kinetic energy acquired at B
• This kinetic energy will be used up by doing work against friction
2. Let the distance moved after 'B' be 'x' m
• Then work done by friction = Frictional force × distance = 5x
3. This much energy will be extracted from the object
• Thus we can write: 95.75 = 5x
• So x = 19.15 m
Another method:
1. Consider the motion beyond 'B'
• The initial velocity v0 for this motion is 2.526
• The final velocity v for this motion is 0 ms-1
2. During this motion, the only force acting on the body is the frictional force which is equal to -5 N
• So acceleration = Force⁄mass = -5⁄30 = -1⁄6 ms-2
3. We can use the equation: $\mathbf\small{v^2-v_0^2=2ax}$
• Entering known values, we get: $\mathbf\small{0^2-2.526^2=2 \times \frac{-1}{6} \times x}$
• From this we get: x = 19.142 m
• This result is same as the one we obtained earlier
Now we will discuss about the effect of varying force
First we will see a constant force. We will write the steps:
1. Consider the graph shown in fig.6.10(a) below
 |
| Fig.6.10 |
• Force F is plotted along the y axis
• So it is a Force-Displacement graph
2. Consider the cyan horizontal line
• It intersects the y axis at P. The y coordinate of P is F1
• So the cyan horizontal line indicates a constant force F1 acting on a body
3. Under the action of the constant force F1, the body is being continuously displaced
• Assume that the path along which the body is displaced is the x axis
4. On the x axis (that is., on the path of the body), two points A and B are marked
• The x coordinate of A is x1
• The x coordinate of B is x2
• Vertical red lines are drawn at A and B
5. Consider the ‘time interval in which the body traveled from A to B’
• We want the kinetic energy acquired by the body during this time interval
• That is., we want the work done on the body during the travel from A to B
6. We know that, work done is (force × displacement)
• So in our present case the work done from A to B is (F1 × ‘distance AB’)
7. Now consider fig.6.10(b) above
• It is the same graph in fig.a. The only difference is that, a particular area is shaded
8. We can see that the shaded area is enclosed between two sets of lines:
• First set of lines: Force graph and the x axis on top and bottom
• Second set of lines: Vertical red lines on the left and right
9. This shaded area is a rectangle
• The length of the rectangle is the ‘distance AB’
• Height of the rectangle is F1
• So area of the rectangle is (F1 × ‘distance AB’)
10. Comparing (6) and (9), we obtain the following relation:
■ Work done from A to B = Area enclosed between A and B
Now we will see a varying force.
1. In real life situations, the forces applied are not constant.
• That means, we do not get a perfect horizontal cyan line as in fig.6.10(a)
• The graph of the force will be a curve as shown in fig.6.11(a) below:
 |
| Fig.6.11 |
• But overall, it is a curve, indicating that it is a varying force
3. The area enclosed by the curve is shown in fig.6.11(b)
• We want to prove that, this area shown in fig.6.11(b) is equal to the work done (from A to B) by the varying force.
4. Before proving that, we must device a method to find the area of such irregular figures. Let us try:
• The same area in fig.6.11(b) is shown in fig.6.12(a) below:
 |
| Fig.6.12 |
5. Those strips have two important features:
(i) Each strip is a perfect rectangle
(ii) The widths of all the strips are the same
6. Since any one strip is a perfect rectangle, we can calculate it’s area easily. Let us see how:
• If we divide the ‘length AB’ by 6, we will get the width of strip
• All strips have the same width. So we can call it the ‘common width’
• This ‘common width’ represents the ‘change in position x’
• So we can indicate the ‘common width’ by 'Δx' (Delta x)
■ We can say: All the strips have the same width Δx
7. Now to find the height of any one strip, look at it’s top-left corner
• That corner touches the curve
• So force ‘F’ at the ‘touching point’ of a rectangle can be taken as it’s height
8. The ‘touching point’ is different for different rectangles.
• Each rectangle has it’s own ‘unique value of F at the touching point’
• We can write:
♦ The value of F at the touching point of first rectangle is F1
♦ The value of F at the touching point of second rectangle is F2
♦ The value of F at the touching point of third rectangle is F3
♦ - - -
♦ - - -
♦ The value of F at the touching point of sixth rectangle is F6
9. Once we know the height of a rectangle, we can multiply it by the ‘common width Δx’ to calculate it’s area
• Thus we can calculate the 6 areas: A1, A2, A3, . . . , A6
♦ A1 = F1 × Δx
♦ A2 = F2 × Δx
♦ A3 = F3 × Δx
♦ - - -
♦ - - -
♦ A6 = F6 × Δx
10. Once we calculate all the 6 areas, we add them
• That is., we find the sum: A1 + A2 + A3 + . . . + A6
11. If all the areas were the same, we could have just written: Sum = 6A
• But here the areas are different
• We use a mathematical notation to express this type of summation
■ This notation is called ‘sigma notation’.
• It’s symbol is the Greek capital letter ($\mathbf\small{\Sigma}$)
• We write 'Sum of the 6 areas' as: $\mathbf\small{\sum_{i=1}^{i=6} A_i}$
11. Let us analyze the above notation:
• The ‘$\mathbf\small{A_i}$’ on the right side of ‘=’ sign conveys this information:
Certain ‘areas’ are to be added together
• How many areas are to be added?
• The answer is given by the ‘i=1’ and ‘i=6’ on bottom and top of $\mathbf\small{\Sigma}$
♦ ‘i= 1’ indicates A1
♦ ‘i= 6’ indicates A6
12. Thus we can write: $\mathbf\small{\sum_{i=1}^{i=6} A_i=A_1+A_2+\;.\;.\;.\;+A_6}$
• In the above expression, the right side is already understood and hence not written
■ So we can write:
The total area of the 6 rectangles in fig.6.12(a) is $\mathbf\small{\sum_{i=1}^{i=6} A_i}$
• We can expand $\mathbf\small{A_i}$.
♦ Expanding is possible because $\mathbf\small{A_i=(F_i \times \Delta x)}$
• We get:
The total area of the 6 rectangles in fig.6.12(a) = $\mathbf\small{\sum_{i=1}^{i=6} (F_i \times \Delta x)}$
13. So we successfully calculated the total area of the six rectangles
• But is this ‘total area’ equal to the shaded area in fig.6.11(b)?
■ No. It isn’t.
14. Let us see the reason:
• Looking closely at fig.6.12(a) we see these two facts:
(i) At the rising part of the cyan curve on the left, there are ‘black triangles’ just below the curve
(ii) At the falling part of the cyan curve on the right, there are ‘colored triangles’ just above the curve
• We have to add the areas of those 'black triangles' to the area in (12)
• Also we have to subtract the areas of those 'colored triangles' from the area in (12)
• Then only we will get a ‘perfect fit’ and thus a correct ‘total area’
16. Another important point can also be noted at this stage:
• Consider any one of the black triangles. One such triangle is shown enlarged in fig.6.12(b)
• It is named as triangle UVW
♦ U is the top-left corner of the rectangle. It is the 'touching point'
♦ V is the top-right corner of the rectangle
♦ W is the top-left corner of the adjacent rectangle
• We used the 'value of F' at the 'touching point' U as the height of that particular rectangle
• Note that, at the left point 'U' of the rectangle, the force acting is the 'force at U'
♦ We will indicate this as Fu
• But at the right point 'V', the force acting is the 'force at W'
♦ We will indicate this as Fw
• That means, during the displacement (Δx) from the bottom-left corner to bottom-right side of this rectangle, the force varies from Fu to Fw
■ That means, during the displacement (Δx) of this rectangle, the force is not constant.
• We will have to apply this information later in this discussion
17. Now to continue with our main discussion:
■ We need to improve our method so that:
Total area of the strips (rectangles) = Shaded area in fig.6.11(b)
• Let us try:
In fig.6.13(a) below, the same cyan curve is shown. The length AB is also the same
 |
| Fig.6.13 |
• So the common width Δx will be less than the 'Δx in fig.6.12(a)'
18. The total area of the 9 rectangles is $\mathbf\small{\sum_{i=1}^{i=9} (F_i \times \Delta x)}$
• Here also we see black triangles and colored triangles
• So we can write:
Total area of the 9 strips (rectangles) ≠ Shaded area in fig.6.11(b)
19. But there is an improvement:
• The triangles have become smaller
• That means accuracy have increased. But not to the expected level
20. Let us try to improve further:
In fig.6.13(b) above, the same cyan curve is shown. The length AB is also the same
• But AB is now divided into 14 strips
• So the common width Δx will be less than the 'Δx in fig.6.13(a)'
21. The total area of the 14 rectangles is $\mathbf\small{\sum_{i=1}^{i=14} (F_i \times \Delta x)}$
• Here also we see black triangles and colored triangles
• So we can write:
Total area of the 14 strips (rectangles) ≠ Shaded area in fig.6.11(b)
22. But there is an improvement:
• The triangles have become still smaller
• That means accuracy have increased. But not to the expected level
23. Let us try to improve further:
In fig.6.14(a) below, the same cyan curve is shown. The length AB is also the same
 |
| Fig.6.14 |
• So the common width Δx will be less than the 'Δx in fig.6.13(b)'
21. The total area of the 19 rectangles is $\mathbf\small{\sum_{i=1}^{i=19} (F_i \times \Delta x)}$
• Here also we see black triangles and colored triangles
• So we can write:
Total area of the 19 strips (rectangles) ≠ Shaded area in fig.6.11(b)
22. But there is much improvement:
• The triangles have become very small
• That means accuracy have increased very much. But not to the expected level
23. So we tried 6 strips, 9 strips, 14 strips and 19 strips.
24. What if we make the 'number of strips' equal to infinity ($\mathbf\small{\infty}$)?
• Then the width of each strip will be $\mathbf\small{\frac{AB}{\infty}}$
• But any thing divided by $\mathbf\small{\infty}$ is zero.
• So $\mathbf\small{\frac{AB}{\infty}=0}$
25. That means, if we increase the 'number of strips' to $\mathbf\small{\infty}$, the widths of the strips will become zero
• Strips of 'zero width' are not helpful for calculating total area
26. So we need to prevent the widths from becoming zero
• And at the same time we need a very large number of strips also
• This situation is represented as: Total area of strips = $\mathbf\small{\lim_{i\to \infty}\sum A_i}$
27. Let us analyse the above expression:
• $\mathbf\small{\sum A_i}$ indicates that 'sum of areas' is to be calculated
• How many areas are there?
The answer is given by $\mathbf\small{\lim_{i\to \infty}}$
• 'i' tends to become infinity.
♦ If it becomes infinity, there is no use because strips will then have zero widths
• So 'i' is kept within a limit (indicated by 'lim')
• It indicates that 'i' should make a 'maximum allowable closeness' to infinity
28. In such a situation, the width of strips will have 'least possible values' like 0.00001 mm
• There can be any number of zeros between the 'decimal point' and '1'
• There is no restriction on that 'number of zeroes'
29. Now we can expand the expression in (26). We get:
• Total area of strips = $\mathbf\small{\lim_{\Delta x \,\to \,0}\sum_{x=x_i}^{x=x_f} (F_x \times \Delta x)}$
30. Let us analyze the above expression:
• $\mathbf\small{\sum (F_x \times \Delta x)}$ indicates that the following sum is to be calculated:
$\mathbf\small{(F_1 \times \Delta x)+(F_2 \times \Delta x)+(F_3 \times \Delta x)+\;.\;.\;.}$
• How many terms are there?
'That many terms' by which, the width Δx of each strip has 'a value very close to zero'
♦ This is indicated by: $\mathbf\small{\lim_{\Delta x \,\to \,0}}$
• The height of each strip is given by the force at the 'touching point'. It is denoted as $\mathbf\small{(F_x}$
♦ $\mathbf\small{F_{xi}}$ is the height of the first strip
♦ $\mathbf\small{F_{xf}}$ is the height of the last strip
31. We saw that, when the number of strips is increased, the triangles will become smaller and smaller
• Finally, when the number of strips approach infinity, those triangles will completely vanish
♦ This is shown in fig.6.14(b)
• That means, the area obtained in (29) is the required shaded area in fig.6.11(b)
32. Now we take up the point that was mentioned in (16)
• When the width of strips approach zero, the triangles vanish
• As shown in fig.6.14(b), the points U, V and W nearly coincide
• Then Fu is same as Fw.
• That means, the same force is acting at the left side and right side of the rectangle
33. This is applicable to all rectangles. We can write the following steps:
(i) Take any one rectangle
• Since Δx is so small, the force which displace the object from the bottom-left corner to the bottom-right corner of that rectangle is a constant
(ii) Since the force is constant for that rectangle, the work done on the object while displacing from the bottom-left corner to the bottom-right corner of that rectangle is given by: (Force corresponding to that rectangle × Δx)
(iii) But (Force corresponding to that rectangle × Δx) = Area of that rectangle
(iv) So 'sum of areas of all rectangles' is the total work done from A to B
34. 'Sum of areas of all rectangles' is given by (29). Thus we can write:
Sum of areas of all rectangles
= Area of the shaded portion in fig.6.11(b)
= Work done from A to B
= $\mathbf\small{\lim_{\Delta x \,\to \,0}\sum_{x=x_i}^{x=x_f} (F_x \times \Delta x)}$
35. So we need to add a very large number of rectangular areas
• But there is nothing to worry about. Such a summation can be done very easily using principles of calculus
• So we will revisit this topic after completing the basic math course on calculus
36. The body is moving along the x axis
• At the point 'A', it has a certain amount of kinetic energy
♦ It is given by: $\mathbf\small{K_A=\frac{1}{2}m|\vec{v_A}|^2}$
• Since the body is acted upon by a force, it will experience acceleration
• So it's velocity at B will be different. Consequently, the kinetic energy at B will also be different
♦ We can write: $\mathbf\small{K_B=\frac{1}{2}m|\vec{v_B}|^2}$
• According to the work-energy theorem, the change in kinetic energy is equal to the work done on the body from A to B
♦ But this work is equal to the area enclosed by the curve.
■ So we can write:
According to the work-energy theorem, the change in kinetic energy is equal to the area enclosed by the curve for a varying force
37. We will see the proof after completing the basic math course on calculus
At this stage, all we need to know are the following two points:
(i) Area enclosed by the 'curve representing a force' is equal to the work done by that force
• For a constant force, we proved this using fig.6.10 above
• For a varying force, we proved this using figs.6.11 to 6.14 above
(ii) 'Change in kinetic energy' acquired by the body is equal to the 'work done by the force' on that body
• For a constant force, we proved this in the previous section, using the equation $\mathbf\small{v^2-v_0^2=2ax}$
• For a varying force, we will see the proof after learning calculus
No comments:
Post a Comment