In the previous section, we saw the law of conservation of energy. In this section, we will see the energy associated with a spring.
1. Consider a spring whose one end is attached to the ceiling. Fig.6.28(a) below:
• A red dashed horizontal line is drawn through the lower end of the spring
• This red dashed line is important for our calculations
2. In fig.b, a mass of 5 kg is attached to the lower end
• We can see that, now the lower end of the spring is 2 cm below the red line
3. The lower end of the spring coincided with the red line when no load was attached
• Now it is 2 cm below
■ So it is clear that, the spring is stretched by a displacement of 2 cm
• We denote this displacement as 'x'
4. We know that, for stretching or compressing a spring, force is required
• Here we have certainly applied a stretching force of (5 kg × g)
• Taking g as 10 ms-2, we get: Stretching force F = 50 N
• We can write: When force F = 50 N, the displacement x of the spring = 2 cm
• Since we have to repeat the experiment, we will give special notations:
F1 = 50 N, x1 = 2 cm
5. Now we repeat the experiment
• We remove the 5 kg mass and attach a 12 kg mass. This is shown in fig.c
• We can see that, now the lower end of the spring is 4.8 cm below the red line
• The lower end of the spring coincided with the red line when no load was attached
• Now it is 4.8 cm below
• So it is clear that, the spring is stretched by a displacement (x) of 4.8 cm
• We can write: F2 = 120 N, x2 = 4.8 cm
6. We repeat the experiment one more time
• We remove the 12 kg mass and attach a 8 kg mass. This is shown in fig.d
• We can see that, now the lower end of the spring is 4.8 cm below the red line
• The lower end of the spring coincided with the red line when no load was attached
• Now it is 4.8 cm below
• So it is clear that, the spring is stretched by a displacement (x) of 4.8 cm
• We can write: F3 = 120 N, x3 = 3.2 cm
7. Now we find the ratio $\mathbf\small{\frac{F}{x}}$ for each case. We get:
• Experiment 1: = $\mathbf\small{\frac{F_1}{x_1}=\frac{50}{2}=25}$
• Experiment 2: = $\mathbf\small{\frac{F_2}{x_2}=\frac{120}{4.8}=25}$
• Experiment 3: = $\mathbf\small{\frac{F_3}{x_3}=\frac{80}{3.2}=25}$
8. We find that, the ratio is a constant for all combinations
• We can repeat the experiment for any number of times we like. We will get the same value of 25
• Since $\mathbf\small{\frac{F}{x}}$ is constant for a spring, it is give a special name: Spring constant
• It is denoted by k
• F in the numerator has unit N and x in the denominator has unit m. So the unit of 'k' is Nm-1.
8. So we have $\mathbf\small{\frac{F}{x}=k}$
• If 'k' is very large, the left side $\mathbf\small{\frac{F}{x}}$ must also be very large
• For $\mathbf\small{\frac{F}{x}}$ to be large, the numerator F should be very large
♦ and denominator x should be very small
9. That means, if 'k' is very large, force F also will have to be very large even to produce a small x
■ Such a spring with large 'k' value is called a stiff spring
• If 'k' is small, F need not be so large to produce significant x
■ Such a spring with small 'k' value is called a soft spring.
■ Every spring has it's own unique value of 'k'
10. Rearranging $\mathbf\small{\frac{F}{x}=k}$, we get: $\mathbf\small{F=kx}$
• Fom this expression, it is clear that, x is proportional to F
• The above properties of spring were first discovered in 1660 by the British scientist Robert Hooke
• It is known by his name as: Hooke's Law
• In his book, he wrote: “as the extension, so the force”. That means, extension is proportional to the force.
• This proportionality is clear from the equation F= kx.
♦ When F increases, x also increases.
♦ When F decreases, x also decreases
(Some basic lessons on proportionality can be seen here)
• However, Hookes' law is stated in the form: F = -kx
• We will now see the reason for the negative sign:
We will write the steps:
1. In fig,6.29 (a) below, a block of mass m rests on a smooth horizontal plane
• It is attached to one end of a spring
• The other end of the spring is rigidly attached to a vertical wall
2. In fig.a, the spring is in it’s normal configuration. That is., it is neither stretched nor compressed
• In this configuration, a red dashed vertical line is drawn through the 'point of contact' between the block and the spring
3. Since at this position, the spring is neither stretched or compressed, the displacement x is zero
• So we can write x = 0 along this dashed line
4. In fig.6.29(b), the block is pulled towards the right by a distance x
• So the spring is stretched by a distance x
5. To keep the spring in this stretched position, a force need to be applied
• We know that, this force is given by F = kx
• If we know the value of the spring constant k, we can multiply that k by the displacement x to obtain F
• So magnitude of F required to keep the block at the position in fig.b can be easily obtained
6. Now we will see the direction of F:
• When the spring is stretched by an external force F, an internal force (spring force Fs) of the same magnitude will develop inside the spring
• This Fs will try to restore it’s normal configuration
• That means, Fs will be opposite in direction to F
7. We applied F towards the right. That is., towards the positive side of the x axis
• So F can be considered to be positive
■ Then internal force Fs will be negative
8. Now we will consider compression. In fig.c, the block is pushed towards the left by a distance x
• So the spring is compressed by a distance x
9. To keep the spring in this compressed position, a force need to be applied
• We know that, this force is given by F = kx
• If we know the value of the spring constant k, we can multiply it with the displacement x to obtain F
• So magnitude of F required to keep the block at the position in fig.c can be easily obtained
10. Now we will see the direction of F:
• When the spring is compressed by an external force F, an internal force (spring force Fs) of the same magnitude will develop inside the spring
• This Fs will try to restore it’s normal configuration
• That means, Fs will be opposite in direction to F
11. We applied F towards the left. That is., towards the negative side of the x axis
• So F can be considered to be negative
■ Then internal force Fs will be positive
12. So we have a situation in which two conditions are involved:
(i) From (7) we have:
When x is positive, Fs is negative
(ii) From (11) we have:
When x is negative, Fs is positive
■ To satisfy both the conditions, we must write: Fs = -kx
We will write the steps:
1. In the relation Fs = -kx:
• The left side gives the internal force developed within the spring
• We enter the displacement x on the right side
2. When x changes, Fs will also change
• So this relation can be used to plot the Force-Displacement graph
• What will be the shape of this graph?
3. In the case of gravitational force, the force does not change with 'distance from the ground'
• Remember that, we plot displacement along x axis and force along y axis
• So the force-displacement graph in the case of gravity will be a horizontal line
4. But the spring force varies with distance
• Luckily for us, it is a uniform variation. Because, force is proportional to displacement
• For uniform variation, the graph will be a straight line
♦ But this straight line will be neither horizontal nor vertical
♦ It will be inclined
(If it is not a uniform variation, the graph will be a curve)
5.So our present graph showing the relation Fs = -kx will be an inclined straight line
• Let us see it’s peculiarities:
(i) From analytic geometry classes, we know that the equation of a straight line is y = mx +c
• If the constant c = 0, then y intercept will be zero
• That means, the straight line will then pass through the origin
• Further, if the slope m is negative, the line will slope downwards (with increase in x)
(ii) So Fs = -kx is the equation of a line which pass through the origin
• Also, with the increase in x, it will be sloping downwards
• This is shown by the cyan line in fig.d
1. We know that work done by a force is equal to the ‘area enclosed by the force-displacement graph. (Details here)
2. Let the spring be stretched by a distance xt as shown in the fig.6.30(a) below
• Then the work done by Fs will be equal to the area of the red triangle shown in the fig.6.30(b)
3. It is easy to calculate this area. We have:
• Base OP of the red triangle = xt
• Altitude PQ of the red triangle = Magnitude of Fs at xt
♦ The relation Fs = kx gives the magnitude of Fs at any x
♦ So Fs at xt = k×xt
■ So area = 1⁄2 × base × altitude = $\mathbf\small{\frac{1}{2}\times x_t \times k \times x_t=\frac{k\,x_t^2}{2}}$
■ Thus we can write:
Work done by the spring force Fs when the spring is stretched through a distance of xt = $\mathbf\small{\frac{k\,x_t^2}{2}}$
4. When we discussed about gravitational potential energy (Pg), we saw the following 7 points:
(i) Gravitational force is the original force. With out that, there is no Pg
(ii) So work done by gravity is taken as positive
(iii) When an object is raised to a height by an external force, work is done against gravity.
(iv) So work done by the external force is taken as negative
(v) That means., the Pg stored in an object (of mass m) at a height h = -mgh joules
(vi) When the object falls, work is done by gravity
(vii) So the amount of energy released when the object is in free fall is +mgh joules
5. We can follow the same sign convention here also
• Fs and the external force F are numerically equal. So work done by Fs is numerically equal to the work done by external force F
• Since F is opposite to the original force Fs, we can write:
■ Work done by F when the spring is stretched (from normal configuration) by xt = $\mathbf\small{-\frac{k\,x_t^2}{2}}$
■ Work done by Fs when the spring is stretched (from normal configuration) by xt = $\mathbf\small{\frac{k\,x_t^2}{2}}$
• This much work is stored in the spring as ‘spring potential energy’
1. We know that work done by a force is equal to the ‘area enclosed by the force-displacement graph
2. Let the spring be compressed by a distance xc as shown in the fig.6.31(a) below
• Then the work done by Fs will be equal to the area of the magenta triangle shown in the fig.6.31(b)
3. It is easy to calculate this area. We have:
• Base OR of the magenta triangle = xc
• Altitude RS of the magenta triangle = Magnitude of Fs at xc
♦ The relation Fs = kx gives the magnitude of Fs at any x
♦ So Fs at xc = k×xc
■ So area = 1⁄2 × base × altitude = $\mathbf\small{\frac{1}{2}\times x_c \times k \times x_c=\frac{k\,x_c^2}{2}}$
■ Thus we can write:
Work done by the spring force Fs when the spring is compressed through a distance of xc = $\mathbf\small{\frac{k\,x_c^2}{2}}$
4. We can follow the same sign convention as the gravitational potential energy, here also
• Fs and the external force F are numerically equal. So work done by Fs is numerically equal to the work done by external force F
• Since F is opposite to the original force Fs, we can write:
■ Work done by F when the spring is compressed (from normal configuration) by xc = $\mathbf\small{-\frac{k\,x_c^2}{2}}$
■ Work done by Fs when the spring is compressed (from normal configuration) by xc = $\mathbf\small{\frac{k\,x_c^2}{2}}$
• This much work is stored in the spring as ‘spring potential energy’
1. Consider a spring whose one end is attached to the ceiling. Fig.6.28(a) below:
 |
| Fig.6.28 |
• This red dashed line is important for our calculations
2. In fig.b, a mass of 5 kg is attached to the lower end
• We can see that, now the lower end of the spring is 2 cm below the red line
3. The lower end of the spring coincided with the red line when no load was attached
• Now it is 2 cm below
■ So it is clear that, the spring is stretched by a displacement of 2 cm
• We denote this displacement as 'x'
4. We know that, for stretching or compressing a spring, force is required
• Here we have certainly applied a stretching force of (5 kg × g)
• Taking g as 10 ms-2, we get: Stretching force F = 50 N
• We can write: When force F = 50 N, the displacement x of the spring = 2 cm
• Since we have to repeat the experiment, we will give special notations:
F1 = 50 N, x1 = 2 cm
5. Now we repeat the experiment
• We remove the 5 kg mass and attach a 12 kg mass. This is shown in fig.c
• We can see that, now the lower end of the spring is 4.8 cm below the red line
• The lower end of the spring coincided with the red line when no load was attached
• Now it is 4.8 cm below
• So it is clear that, the spring is stretched by a displacement (x) of 4.8 cm
• We can write: F2 = 120 N, x2 = 4.8 cm
6. We repeat the experiment one more time
• We remove the 12 kg mass and attach a 8 kg mass. This is shown in fig.d
• We can see that, now the lower end of the spring is 4.8 cm below the red line
• The lower end of the spring coincided with the red line when no load was attached
• Now it is 4.8 cm below
• So it is clear that, the spring is stretched by a displacement (x) of 4.8 cm
• We can write: F3 = 120 N, x3 = 3.2 cm
7. Now we find the ratio $\mathbf\small{\frac{F}{x}}$ for each case. We get:
• Experiment 1: = $\mathbf\small{\frac{F_1}{x_1}=\frac{50}{2}=25}$
• Experiment 2: = $\mathbf\small{\frac{F_2}{x_2}=\frac{120}{4.8}=25}$
• Experiment 3: = $\mathbf\small{\frac{F_3}{x_3}=\frac{80}{3.2}=25}$
8. We find that, the ratio is a constant for all combinations
• We can repeat the experiment for any number of times we like. We will get the same value of 25
• Since $\mathbf\small{\frac{F}{x}}$ is constant for a spring, it is give a special name: Spring constant
• It is denoted by k
• F in the numerator has unit N and x in the denominator has unit m. So the unit of 'k' is Nm-1.
8. So we have $\mathbf\small{\frac{F}{x}=k}$
• If 'k' is very large, the left side $\mathbf\small{\frac{F}{x}}$ must also be very large
• For $\mathbf\small{\frac{F}{x}}$ to be large, the numerator F should be very large
♦ and denominator x should be very small
9. That means, if 'k' is very large, force F also will have to be very large even to produce a small x
■ Such a spring with large 'k' value is called a stiff spring
• If 'k' is small, F need not be so large to produce significant x
■ Such a spring with small 'k' value is called a soft spring.
■ Every spring has it's own unique value of 'k'
10. Rearranging $\mathbf\small{\frac{F}{x}=k}$, we get: $\mathbf\small{F=kx}$
• Fom this expression, it is clear that, x is proportional to F
• The above properties of spring were first discovered in 1660 by the British scientist Robert Hooke
• It is known by his name as: Hooke's Law
• In his book, he wrote: “as the extension, so the force”. That means, extension is proportional to the force.
• This proportionality is clear from the equation F= kx.
♦ When F increases, x also increases.
♦ When F decreases, x also decreases
(Some basic lessons on proportionality can be seen here)
• However, Hookes' law is stated in the form: F = -kx
• We will now see the reason for the negative sign:
We will write the steps:
1. In fig,6.29 (a) below, a block of mass m rests on a smooth horizontal plane
 |
| Fig.6.29 |
• The other end of the spring is rigidly attached to a vertical wall
2. In fig.a, the spring is in it’s normal configuration. That is., it is neither stretched nor compressed
• In this configuration, a red dashed vertical line is drawn through the 'point of contact' between the block and the spring
3. Since at this position, the spring is neither stretched or compressed, the displacement x is zero
• So we can write x = 0 along this dashed line
4. In fig.6.29(b), the block is pulled towards the right by a distance x
• So the spring is stretched by a distance x
5. To keep the spring in this stretched position, a force need to be applied
• We know that, this force is given by F = kx
• If we know the value of the spring constant k, we can multiply that k by the displacement x to obtain F
• So magnitude of F required to keep the block at the position in fig.b can be easily obtained
6. Now we will see the direction of F:
• When the spring is stretched by an external force F, an internal force (spring force Fs) of the same magnitude will develop inside the spring
• This Fs will try to restore it’s normal configuration
• That means, Fs will be opposite in direction to F
7. We applied F towards the right. That is., towards the positive side of the x axis
• So F can be considered to be positive
■ Then internal force Fs will be negative
8. Now we will consider compression. In fig.c, the block is pushed towards the left by a distance x
• So the spring is compressed by a distance x
9. To keep the spring in this compressed position, a force need to be applied
• We know that, this force is given by F = kx
• If we know the value of the spring constant k, we can multiply it with the displacement x to obtain F
• So magnitude of F required to keep the block at the position in fig.c can be easily obtained
10. Now we will see the direction of F:
• When the spring is compressed by an external force F, an internal force (spring force Fs) of the same magnitude will develop inside the spring
• This Fs will try to restore it’s normal configuration
• That means, Fs will be opposite in direction to F
11. We applied F towards the left. That is., towards the negative side of the x axis
• So F can be considered to be negative
■ Then internal force Fs will be positive
12. So we have a situation in which two conditions are involved:
(i) From (7) we have:
When x is positive, Fs is negative
(ii) From (11) we have:
When x is negative, Fs is positive
■ To satisfy both the conditions, we must write: Fs = -kx
Now we know the reason for the negative sign. Next we will see the peculiarities of this relation
1. In the relation Fs = -kx:
• The left side gives the internal force developed within the spring
• We enter the displacement x on the right side
2. When x changes, Fs will also change
• So this relation can be used to plot the Force-Displacement graph
• What will be the shape of this graph?
3. In the case of gravitational force, the force does not change with 'distance from the ground'
• Remember that, we plot displacement along x axis and force along y axis
• So the force-displacement graph in the case of gravity will be a horizontal line
4. But the spring force varies with distance
• Luckily for us, it is a uniform variation. Because, force is proportional to displacement
• For uniform variation, the graph will be a straight line
♦ But this straight line will be neither horizontal nor vertical
♦ It will be inclined
(If it is not a uniform variation, the graph will be a curve)
5.So our present graph showing the relation Fs = -kx will be an inclined straight line
• Let us see it’s peculiarities:
(i) From analytic geometry classes, we know that the equation of a straight line is y = mx +c
• If the constant c = 0, then y intercept will be zero
• That means, the straight line will then pass through the origin
• Further, if the slope m is negative, the line will slope downwards (with increase in x)
(ii) So Fs = -kx is the equation of a line which pass through the origin
• Also, with the increase in x, it will be sloping downwards
• This is shown by the cyan line in fig.d
Next we will find the work done by Fs when the spring is stretched :
2. Let the spring be stretched by a distance xt as shown in the fig.6.30(a) below
 |
| Fig.6.30 |
3. It is easy to calculate this area. We have:
• Base OP of the red triangle = xt
• Altitude PQ of the red triangle = Magnitude of Fs at xt
♦ The relation Fs = kx gives the magnitude of Fs at any x
♦ So Fs at xt = k×xt
■ So area = 1⁄2 × base × altitude = $\mathbf\small{\frac{1}{2}\times x_t \times k \times x_t=\frac{k\,x_t^2}{2}}$
■ Thus we can write:
Work done by the spring force Fs when the spring is stretched through a distance of xt = $\mathbf\small{\frac{k\,x_t^2}{2}}$
4. When we discussed about gravitational potential energy (Pg), we saw the following 7 points:
(i) Gravitational force is the original force. With out that, there is no Pg
(ii) So work done by gravity is taken as positive
(iii) When an object is raised to a height by an external force, work is done against gravity.
(iv) So work done by the external force is taken as negative
(v) That means., the Pg stored in an object (of mass m) at a height h = -mgh joules
(vi) When the object falls, work is done by gravity
(vii) So the amount of energy released when the object is in free fall is +mgh joules
5. We can follow the same sign convention here also
• Fs and the external force F are numerically equal. So work done by Fs is numerically equal to the work done by external force F
• Since F is opposite to the original force Fs, we can write:
■ Work done by F when the spring is stretched (from normal configuration) by xt = $\mathbf\small{-\frac{k\,x_t^2}{2}}$
■ Work done by Fs when the spring is stretched (from normal configuration) by xt = $\mathbf\small{\frac{k\,x_t^2}{2}}$
• This much work is stored in the spring as ‘spring potential energy’
Next we will find the work done by Fs when the spring is compressed:
2. Let the spring be compressed by a distance xc as shown in the fig.6.31(a) below
 |
| Fig.6.31 |
3. It is easy to calculate this area. We have:
• Base OR of the magenta triangle = xc
• Altitude RS of the magenta triangle = Magnitude of Fs at xc
♦ The relation Fs = kx gives the magnitude of Fs at any x
♦ So Fs at xc = k×xc
■ So area = 1⁄2 × base × altitude = $\mathbf\small{\frac{1}{2}\times x_c \times k \times x_c=\frac{k\,x_c^2}{2}}$
■ Thus we can write:
Work done by the spring force Fs when the spring is compressed through a distance of xc = $\mathbf\small{\frac{k\,x_c^2}{2}}$
4. We can follow the same sign convention as the gravitational potential energy, here also
• Fs and the external force F are numerically equal. So work done by Fs is numerically equal to the work done by external force F
• Since F is opposite to the original force Fs, we can write:
■ Work done by F when the spring is compressed (from normal configuration) by xc = $\mathbf\small{-\frac{k\,x_c^2}{2}}$
■ Work done by Fs when the spring is compressed (from normal configuration) by xc = $\mathbf\small{\frac{k\,x_c^2}{2}}$
• This much work is stored in the spring as ‘spring potential energy’
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