In the previous section, we saw the potential energy when a spring is stretched/extended horizontally. We will now see the potential energy when a spring is stretched/extended vertically.
1. Consider a spring with spring constant 'k'.
• One end of it is attached to the roof. No mass is attached to the other end.
• This is shown in fig.6.38(a) below:
2. The spring is now in it's 'natural length' because, it is neither compressed nor stretched.
• A red horizontal dashed line is drawn through it's bottom end.
• This line will show the equilibrium position. We will call it 'Equilibrium-1'
• It is at a height of h1 from the ground level
[Note that, if the spring is heavy, it will stretch under it's own weight when attached vertically to the roof. Then the red dashed line will not show the 'natural length'. Here, we are assuming that, the spring is light]
3. Now let us attach a block of mass 'm' at the free end.
• This is shown in fig.b.
• The 'process of attaching the block' must be done in a special way. Let us see how:
♦ The block should not be 'just attached' and allowed to lower by itself.
♦ Instead, we must hold the block by hand from it's bottom and attach it
♦ After attaching, we must lower our hand gradually so that the spring is stretched gradually
♦ We will soon obtain the 'lower most point', where the block no longer needs the support from our hand
♦ At that point, the block will be completely supported by the spring
♦ There will be no further downward movement for the block
4. It is a point of equilibrium
• Through that point, a green horizontal dashed line is drawn. We will call it 'Equilibrium-2'
• It is at a height of h2 from the ground level
• This is shown in fig.b
[If the block is 'just attached' and allowed to lower by itself, the 'stretching process' of the spring will take place suddenly and the spring will oscillate. For our present discussion, we must avoid such a situation]
5. Let the vertical distance between Equilibrium-1 and Equilibrium-2 be x1 m
• Then the force in the spring in fig.b will be kx1
6. This force will be acting upwards. It will balance the downward force 'mg' of the block
• So we can write: kx1 = mg
• We will be required to apply this result soon
7. Next we pull the block slowly by an extra distance x2
• Through the new end point, a magenta horizontal dashed line is drawn.
• We will call it 'Stretched-Level'
• This is shown in fig.c
8.When the spring is extended downwards by the extra distance x2, we are doing 'extra work' on it
• This extra work is stored as 'extra potential energy' in the spring
9. Next we let the block go
• When we let go of the block, the spring will contract and the block will move upwards
• Due to the upward 'motion', the block will be having kinetic energy
10. When the block reaches Equilibrium-2, it will be having a large kinetic energy
• This kinetic energy will enable the block to climb past Equilibrium-2
11. We want to know this:
• The value of this kinetic energy when the block just passes Equilibrium-2
12. For that, we will use the Law of conservation of energy
• We will choose two levels:
(i) The stretched-level
(ii) The Equilibrium-2
■ By the Law of conservation of energy, the total energy must be the same at those two levels
13. Let us write the various energies:
(i) Total energy Ei at the stretched-level:
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{k\,(x_1 +x_2)^2}{2}}$
∵Total extension = (x1+x2)
• Gravitational potential energy = $\mathbf\small{mgh_3}$
■ So total energy Ei = $\mathbf\small{\frac{k\,(x_1 +x_2)^2}{2}+mgh_3}$
(ii) Total energy Ef at the Equilibrium-2 level
• Kinetic energy = $\mathbf\small{\frac{m\,v^2}{2}}$
♦ Where v is the velocity of the block when it just passes Equilibrium-2
• Spring potential energy = $\mathbf\small{\frac{k\,(x_1)^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_2}$
■ So total energy Ef = $\mathbf\small{\frac{m\,v^2}{2}+\frac{k\,x_1^2}{2}+mgh_2}$
14. Equating the two energies we get:
$\mathbf\small{\frac{k\,(x_1 +x_2)^2}{2}+mgh_3=\frac{m\,v^2}{2}+\frac{k\,x_1^2}{2}+mgh_2}$
15. We can simplify this equation in steps:
• The first step is to bring $\mathbf\small{mgh_3}$ to the right side. We get:
$\mathbf\small{\frac{k\,(x_1 +x_2)^2}{2}=\frac{m\,v^2}{2}+\frac{k\,x_1^2}{2}+mgh_2-mgh_3}$
$\mathbf\small{\Longrightarrow \frac{k\,(x_1 +x_2)^2}{2}=\frac{m\,v^2}{2}+\frac{k\,x_1^2}{2}+mg(h_2-h_3)}$
$\mathbf\small{\Longrightarrow \frac{k\,(x_1 +x_2)^2}{2}=\frac{m\,v^2}{2}+\frac{k\,x_1^2}{2}+mgx_2}$
• Now we multiply both sides by 2. We get:
$\mathbf\small{{k\,(x_1 +x_2)^2}={m\,v^2}+{k\,x_1^2}+2mgx_2}$
• Expanding the left side, we get:
$\mathbf\small{{kx_1^2 + 2kx_1 x_2+kx_2^2}={m\,v^2}+{k\,x_1^2}+2mgx_2}$
• The first term on left side is same as the second term on the right side. So they will cancel each other. We get:
$\mathbf\small{{2kx_1 x_2+kx_2^2}={m\,v^2}+2mgx_2}$
• The first term on the left side is $\mathbf\small{{2kx_1 x_2}}$
• But from (6), $\mathbf\small{kx_1=mg}$
• Thus we get: $\mathbf\small{{2mg x_2+kx_2^2}={m\,v^2}+2mgx_2}$
• Now, the first term on left side is same as the last term on the right side
16. Thus we get: $\mathbf\small{{k\,x_2^2}={m\,v^2}}$
17. From this we get: $\mathbf\small{v=\left [\sqrt{\frac{k}{m}} \right ]x_2}$
• This is the answer to our query in step (11)
■ Note: In the above discussion, h1, h2 and h3 are unknowns. But they do not create any problem while solving. This is because, it is the 'differences between the heights' that matters. We can see that, those differences are known values: x1 and x2. Thus we can say: Any convenient level can be chosen as the datum (ground level) for defining gravitational potential energy.
By applying the Law of conservation of energy, we can solve many problems related to vertical springs. We will see a few solved examples:
Solved example 6.18
A block of mass 0.35 kg is attached to a vertical spring. The spring constant of the spring is 30 Nm-1.
The block is initially supported by hand so that the spring is neither stretched nor compressed. It is then released. How far does the ball fall? [g = 9.8 ms-2]
Solution:
1. Fig.6.39(a) below shows the situation when the block is supported by hand
• This level is marked as 'Equilibrium'
• It is at a height 'h1' from the ground level
• If there is no support from the hand, the block will be at equilibrium at a 'stretched position' further below
2. When the block is released from the equilibrium level in fig.a, it will fall down to a lower level.
• This level is marked as ‘stretched level’ in fig.b
• It is at a height 'h2' from the ground level
• It is at a vertical distance of ‘x’ from the equilibrium
3. The block will remain at the stretched level only momentarily. Because, it will be immediately pulled up by the spring
• We are asked to find the maximum distance traveled downwards from Equilibrium level
• That is., we are asked to find 'x'
4. For that, we will use the Law of conservation of energy
• We will choose two levels:
(i) The Equilibrium
(ii) The stretched-level
■ The total energy must be the same at those two levels
5. Let us write the various energies:
(i) Total energy Ei at the Equilibrium level
• Kinetic energy = 0
• Spring potential energy = 0
♦ ∵ The spring is neither stretched nor compressed
• Gravitational potential energy = $\mathbf\small{mgh_1}$
■ So total energy Ei = $\mathbf\small{mgh_1}$
(ii) Total energy Ef at the stretched-level:
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{k\,x^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_2}$
■ So total energy Ef = $\mathbf\small{\frac{k\,x^2}{2}+mgh_2}$
6. Equating the two energies we get:
$\mathbf\small{mgh_1=\frac{k\,x^2}{2}+mgh_2}$
$\mathbf\small{\Longrightarrow mg(h_1-h_2)=\frac{k\,x^2}{2}}$
$\mathbf\small{\Longrightarrow mgx=\frac{k\,x^2}{2}}$ [∵ (h1-h2) = x]
$\mathbf\small{\Longrightarrow mg=\frac{k\,x}{2}}$
$\mathbf\small{\Longrightarrow x=\frac{2mg}{k}}$
Substituting the known values, we get: x = 0.229 m
■ Note: In this example, h1 and h2 are unknowns. But they do not create any problem while solving. This is because, it is the 'difference between h1 and h2' that matters. We can see that, this difference is 'x'. Thus we can say: Any convenient level can be chosen as the datum (ground level) for defining gravitational potential energy.
Solved example 6.19
A block of mass 5 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below so that it acquires an upward speed of 2 m/s. How high will it rise? [g = 10 ms-2]
Solution:
1. In fig.6.40 below, the natural length of the spring is indicated by 'Equilibrium-1'
• When the block of 5 kg mass is attached, the extension is 10 cm (0.10 m). This is shown in fig.b
2. Using the relation kx = mg, we get:
$\mathbf\small{k=\frac{mg}{x}=\frac{5 \times 10}{0.10}=500\,Nm^{-1}}$
3. For this problem, we need not bother about 'how' or 'why' the impulsive force was given
• All that matters is this:
♦ The block started it's upward journey from Equilibrium-2 with a velocity of 2 ms-1
♦ In other words, the initial velocity of the upward journey is 2 ms-1
4. As a result of this journey, let the block reach the level indicated as 'compressed-level' in fig.c
• Let this 'compressed-level' be at a vertical distance of 'x1' from the Equilibrium level
5. We are asked to find the total distance '(0.1+x1)'
• We can find this total distance once we find x1
6. To find x1, we use the law of conservation of energy
• We will choose two levels:
(i) The Equilibrium-2
(ii) The compressed-level
■ The total energy must be the same at those two levels
7. Let us write the various energies:
(i) Total energy Ei at the Equilibrium-2
• Kinetic energy = $\mathbf\small{\frac{mv^2}{2}=\frac{5 \times 2^2}{2}=10\,J}$
• Spring potential energy = $\mathbf\small{\frac{kx^2}{2}=\frac{500 \times 0.1^2}{2}=2.5\,J}$
• Gravitational potential energy = $\mathbf\small{mgh_2=5 \times 10 \times h_2=50h_2\,\,J}$
■ So total energy Ei = (12.5+50h2) J
(ii) Total energy Ef at the compressed-level:
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{k\,x_1^2}{2}=\frac{500\,x_1^2}{2}=250x_1^2}\,\, J$
• Gravitational potential energy = $\mathbf\small{mgh_3=5 \times 10 \times h_3=50h_3\,\,J}$
■ So total energy Ef = $\mathbf\small{250x_1^2+50h_3}$
8. Equating the two energies we get:
$\mathbf\small{12.5+50h_2=250x_1^2+50h_3}$
$\mathbf\small{\Longrightarrow 12.5=250x_1^2+50(h_3-h_2)}$
$\mathbf\small{\Longrightarrow 12.5=250x_1^2+50(x_1+0.1)}$
• Dividing both sides by 12.5, we get: $\mathbf\small{1=20x_1^2+4(x_1+0.1)}$
$\mathbf\small{\Longrightarrow 1=20x_1^2+4x_1+0.4}$
$\mathbf\small{\Longrightarrow 20x_1^2+4x_1-0.6=0}$
• This is a quadratic equation in x1
• Solving it, we get: x1 = 0.1 or -0.3
• x1 cannot be negative. So we can write: x1 = 0.1 m
■ Thus the answer is: The block rises to a height of 0.2 m above Equilibrium-2
■ Note: In this example, h1, h2 and h3 are unknowns. But they do not create any problem while solving. This is because, it is the 'difference between heights' that matters. Those 'differences' can be easily obtained from the fig. Thus we can say: Any convenient level can be chosen as the datum (ground level) for defining gravitational potential energy.
Solved example 6.20
A vertical spring with constant 200 N/m has a light platform on its top. When a 500 g mass is kept on the platform spring compresses 2.5 cm. Mass is now pushed down 7.50 cm further and released. How far above later position will the mass fly? [g = 10 ms-2]
Solution:
1. In fig.6.41 below, the natural length of the spring is indicated by 'Equilibrium-1'
• When the block of 0.5 kg mass is attached, the compression is 2.5 cm (0.025 m). This is shown in fig.b
2. Using the relation kx = mg, we get:
$\mathbf\small{k=\frac{mg}{x}=\frac{0.5 \times 10}{0.025}=200\,Nm^{-1}}$
3. Now, the block is pushed further down by 0.075 m
• This is shown in fig.c
• The new level is indicated as 'Compressed Level'
4. From this compressed position, the block is released
• The spring will then release it's potential energy and expand
• But it can go only upto the Equilibrium-1 level
• This is shown in fig.d
• The block will have acquired kinetic energy, which will enable it to fly further up above Equilibrium-1
5. The maximum height that the block achieve is shown by a white horizontal dashed line
• We want to find the vertical distance between 'Compressed Level' and 'Maximum Height'
• This vertical distance is marked as x2 in the fig.d
6. To find x2, we use the law of conservation of energy
• We will choose two levels:
(i) The compressed-level
(ii) The Maximum height
■ The total energy must be the same at those two levels
7. Let us write the various energies:
(i) Total energy Ei at the Compressed-level
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{kx^2}{2}=\frac{200 \times 0.1^2}{2}=1.0\,J}$
• Gravitational potential energy = $\mathbf\small{mgh_3=0.5 \times 10 \times h_3=5h_3\,\,J}$
■ So total energy Ei = (1+5h3) J
(ii) Total energy Ef at theMaximum height
• Kinetic energy = 0
• Spring potential energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_4=0.5 \times 10 \times h_4=5h_4\,\,J}$
■ So total energy Ef = $\mathbf\small{5h_4}$
8. Equating the two energies we get:
$\mathbf\small{1+5h_3=5h_4}$
$\mathbf\small{\Longrightarrow 1=5(h_4-h_3)=5x_2}$
$\mathbf\small{\Longrightarrow x_2=\frac{1}{5}=0.2\,m}$
So the block travels a distance of 20 cm from the point of release
Another method:
1. After the block is released, consider the travel from 'Compressed-Level' to 'Equilibrium-1'
■ The total energy must be the same at those two levels
2. Let us write the various energies:
(i) Total energy Ei at the Compressed-level
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{kx^2}{2}=\frac{200 \times 0.1^2}{2}=1.0\,J}$
• Gravitational potential energy = $\mathbf\small{mgh_3=0.5 \times 10 \times h_3=5h_3\,\,J}$
■ So total energy Ei = (1+5h3) J
(ii) Total energy Ef at the Equilibrium-1
• Kinetic energy = $\mathbf\small{\frac{mv^2}{2}=\frac{0.5 \times v^2}{2}=0.25v^2\,J}$
♦ Where 'v' is the velocity acquired by the block when it reaches Equilibrium-1
• Spring potential energy = 0
♦ ∵ The spring is neither stretched nor compressed at Equilibrium-1
• Gravitational potential energy = $\mathbf\small{mgh_1=0.5 \times 10 \times h_1=5h_1\,\,J}$
■ So total energy Ef = $\mathbf\small{0.25v^2+5h_1}$
3. Equating the two energies we get:
$\mathbf\small{1+5h_3=0.25v^2+5h_1}$
$\mathbf\small{\Longrightarrow 1+5(h_3-h_1)=0.25v^2}$
$\mathbf\small{\Longrightarrow 1+5[-(x+x_1)]=0.25v^2}$
$\mathbf\small{\Longrightarrow 1+5[-0.1]=0.25v^2}$
$\mathbf\small{\Longrightarrow 0.5=0.25v^2}$
$\mathbf\small{\Longrightarrow v=\sqrt{2}\,\,ms^{-1}}$
4. Now we have a journey from 'Equilibrium-1' to 'Max. Height'
For this journey:
• The initial velocity is $\mathbf\small{\sqrt{2}\,\,ms^{-1}}$
• Final velocity is zero
• Acceleration = -g = -10 ms-2
5. We can use the third equation of motion:
$\mathbf\small{v^2-u^2=2as}$
• Substituting the known values, we get:
$\mathbf\small{0^2-(\sqrt{2})^2=2 \times -10 \times s}$
$\mathbf\small{\Longrightarrow -2=-20s}$
$\mathbf\small{\Longrightarrow s=0.1\,\,m}$
6. Thus, the distance between ‘Equilibrium-1’ and ‘Maximum height’ is 0.1 m
■ So the required distance between ‘Compressed-level’ and ‘Maximum height’
= 0.1 + (0.025 + 0.075) = 0.2 m (same as above)
■ Note: In this example, h1, h2, h3 and h4 are unknowns. But they do not create any problem while solving. This is because, it is the 'difference between heights' that matters. Those 'differences' can be easily obtained from the fig. Thus we can say: Any convenient level can be chosen as the datum (ground level) for defining gravitational potential energy.
Solved example 6.21
A block of mass 20 kg is released from rest so as to slide in between vertical rails. It compresses a spring (k = 1920 N/m) placed 1 m below the starting point of the block. The rails offer a frictional force of 40 N which opposes the fall of the block. Find (I) The velocity of the block just before striking the spring (ii) The maximum compression of the spring (iii) The distance through which the block is rebounded up, from the maximum compressed position. [g = 10 ms-2]
Solution:
1. In fig.6.42(a) below, the block is at the ‘initial-height’
• The vertical distance between ‘initial-level’ and the top of the spring is 1 m
• The top of the spring is marked as ‘Equilibrium-1’
• We want the velocity of the block just before striking the spring
2. To find that velocity, we use the law of conservation of energy
• We will choose two levels:
(i) The Initial-Height
(ii) The Equilibrium-1
3. Let us write the various energies:
(i) Total energy Ei at the Initial-Height
• Kinetic energy = 0
• Spring potential energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_1=20 \times 10 \times h_1=200h_1\,\,J}$
■ So total energy Ei = (200h1) J
(ii) Total energy Ef at the Equilibrium-1
• Kinetic energy = $\mathbf\small{\frac{mv^2}{2}=\frac{20 \times v^2}{2}=10v^2\,J}$
♦ Where 'v' is the velocity of the block just before striking the spring
• Spring potential energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_0=20 \times 10 \times h_0=200h_0\,\,J}$
■ So total energy Ef = $\mathbf\small{10v^2+200h_0}$
4. By the law of conservation of energy, all the energy at 'Initial-height' must be available at 'Equilibrium-1'
• But some energy is lost for doing work against friction
• Quantity of that 'lost energy' = Frictional force × distance = 40 × 1 = 40 J
5. If we add this 'lost energy' on the right side, then the energies will balance. Thus we get:
$\mathbf\small{200h_1=10v^2+200h_0+40}$
$\mathbf\small{\Longrightarrow 200(h_1-h_0)-40=10v^2}$
$\mathbf\small{\Longrightarrow 200 \times 1-40=10v^2}$
$\mathbf\small{\Longrightarrow 160=10v^2}$
$\mathbf\small{\Longrightarrow v=4\,\,ms^{-1}}$
This is the answer for part (i)
6. The lowest level reached by the block is marked as ‘Compressed-Level’
• We want the vertical height between ‘Equilibrium-1’ and ‘Compressed-level’
7. To find that height, we use the law of conservation of energy
• We will choose two levels:
(i) The Equilibrium-1
(ii) The Compressed-Level
8. Let us write the various energies:
(i) Total energy Ei at the Equilibrium-1
• Kinetic energy = $\mathbf\small{\frac{mv^2}{2}=\frac{20 \times 4^2}{2}=160\,J}$
• Spring potential energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_1=20 \times 10 \times h_1=200h_0\,\,J}$
■ So total energy Ei = (160+200h0) J
(ii) Total energy Ef at the Compressed-Level
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{kx^2}{2}=\frac{1920 \times (h_0-h_2)^2}{2}=960(h_0-h_2)\,J}$
♦ For convenience, let us put (h0-h2) = x1
♦ Then Spring potential energy = 960x12
• Gravitational potential energy = $\mathbf\small{mgh_2=20 \times 10 \times h_2=200h_2\,\,J}$
■ So total energy Ef = 960x12 + 200h2
9. By the law of conservation of energy, all the energy at 'Equilibrium-1' must be available at 'Compressed-Level'
• But some energy is lost for doing work against friction
• Quantity of that 'lost energy' = Frictional force × distance = 40 × x1 J
10. If we add this 'lost energy' on the right side, then the energies will balance. Thus we get:
160+200h0 = 960x12 + 200h2 + 40x1
⇒ 200(h0-h2) = 960x12 + 40x1 - 160
⇒ 200x1 = 960x12 + 40x1 - 160
⇒ 960x12 -160x1 - 160 = 0
• Dividing both sides by 160, we get:
6x12 -x1 - 1 = 0
• This is a quadratic equation in x1. Solving it, we get:
x1 = 0.5 m or -0.333 m
• x1 Cannot be negative. So we can write x1 = 0.5 m
• Thus we get: The maximum compression of the spring = 0.5 m
• This is the answer for part (ii)
11. After reaching the 'compressed-level', the spring will expand
• The block will be pushed upwards and it will fly upto a certain height after leaving the upper end of the spring
• We want the maximum height reached by the block
• This maximum height is marked with white horizontal dashed line
• We want the vertical distance between 'compressed-level' and 'maximum-height'
12. For that, we use the law of conservation of energy
• We will choose two levels:
(i) The compressed-level
(ii) The Maximum height
■ The total energy must be the same at those two levels
13. Let us write the various energies:
(i) Total energy Ei at the Compressed-level
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{kx^2}{2}=\frac{1920 \times 0.5^2}{2}=240\,J}$
• Gravitational potential energy = $\mathbf\small{mgh_3=20 \times 10 \times h_3=200h_2\,\,J}$
■ So total energy Ei = (240+200h2) J
(ii) Total energy Ef at theMaximum height
• Kinetic energy = 0
• Spring potential energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_3=20 \times 10 \times h_3=200h_3\,\,J}$
■ So total energy Ef = $\mathbf\small{200h_3}$
14. By the law of conservation of energy, all the energy at 'compressed-level' must be available at 'maximum-height'
• But some energy is lost for doing work against friction
• Quantity of that 'lost energy' = Frictional force × distance = 40 × (h3-h2)
• For convenience, let us put (h3-h2) = x2
• So 'lost energy' = 40x2
15. If we add this 'lost energy' on the right side, then the energies will balance. Thus we get:
240+200h2 = 200h3 + 40x2
⇒ 240 = 200(h3-h2) + 40x2
⇒ 240 = 200x2 + 40x2
⇒ 240 = 240x2.
⇒ x2 = 1 m
• This is the answer for part (iii)
■ Note: In this example, h0, h1, h2 and h3 are unknowns. But they do not create any problem while solving. This is because, it is the 'difference between heights' that matters. Those 'differences' can be easily obtained from the fig. Thus we can say: Any convenient level can be chosen as the datum (ground level) for defining gravitational potential energy.
1. Consider a spring with spring constant 'k'.
• One end of it is attached to the roof. No mass is attached to the other end.
• This is shown in fig.6.38(a) below:
 |
| Fig.6.38 |
• A red horizontal dashed line is drawn through it's bottom end.
• This line will show the equilibrium position. We will call it 'Equilibrium-1'
• It is at a height of h1 from the ground level
[Note that, if the spring is heavy, it will stretch under it's own weight when attached vertically to the roof. Then the red dashed line will not show the 'natural length'. Here, we are assuming that, the spring is light]
3. Now let us attach a block of mass 'm' at the free end.
• This is shown in fig.b.
• The 'process of attaching the block' must be done in a special way. Let us see how:
♦ The block should not be 'just attached' and allowed to lower by itself.
♦ Instead, we must hold the block by hand from it's bottom and attach it
♦ After attaching, we must lower our hand gradually so that the spring is stretched gradually
♦ We will soon obtain the 'lower most point', where the block no longer needs the support from our hand
♦ At that point, the block will be completely supported by the spring
♦ There will be no further downward movement for the block
4. It is a point of equilibrium
• Through that point, a green horizontal dashed line is drawn. We will call it 'Equilibrium-2'
• It is at a height of h2 from the ground level
• This is shown in fig.b
[If the block is 'just attached' and allowed to lower by itself, the 'stretching process' of the spring will take place suddenly and the spring will oscillate. For our present discussion, we must avoid such a situation]
5. Let the vertical distance between Equilibrium-1 and Equilibrium-2 be x1 m
• Then the force in the spring in fig.b will be kx1
6. This force will be acting upwards. It will balance the downward force 'mg' of the block
• So we can write: kx1 = mg
• We will be required to apply this result soon
7. Next we pull the block slowly by an extra distance x2
• Through the new end point, a magenta horizontal dashed line is drawn.
• We will call it 'Stretched-Level'
• This is shown in fig.c
8.When the spring is extended downwards by the extra distance x2, we are doing 'extra work' on it
• This extra work is stored as 'extra potential energy' in the spring
9. Next we let the block go
• When we let go of the block, the spring will contract and the block will move upwards
• Due to the upward 'motion', the block will be having kinetic energy
10. When the block reaches Equilibrium-2, it will be having a large kinetic energy
• This kinetic energy will enable the block to climb past Equilibrium-2
11. We want to know this:
• The value of this kinetic energy when the block just passes Equilibrium-2
12. For that, we will use the Law of conservation of energy
• We will choose two levels:
(i) The stretched-level
(ii) The Equilibrium-2
■ By the Law of conservation of energy, the total energy must be the same at those two levels
13. Let us write the various energies:
(i) Total energy Ei at the stretched-level:
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{k\,(x_1 +x_2)^2}{2}}$
∵Total extension = (x1+x2)
• Gravitational potential energy = $\mathbf\small{mgh_3}$
■ So total energy Ei = $\mathbf\small{\frac{k\,(x_1 +x_2)^2}{2}+mgh_3}$
(ii) Total energy Ef at the Equilibrium-2 level
• Kinetic energy = $\mathbf\small{\frac{m\,v^2}{2}}$
♦ Where v is the velocity of the block when it just passes Equilibrium-2
• Spring potential energy = $\mathbf\small{\frac{k\,(x_1)^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_2}$
■ So total energy Ef = $\mathbf\small{\frac{m\,v^2}{2}+\frac{k\,x_1^2}{2}+mgh_2}$
14. Equating the two energies we get:
$\mathbf\small{\frac{k\,(x_1 +x_2)^2}{2}+mgh_3=\frac{m\,v^2}{2}+\frac{k\,x_1^2}{2}+mgh_2}$
15. We can simplify this equation in steps:
• The first step is to bring $\mathbf\small{mgh_3}$ to the right side. We get:
$\mathbf\small{\frac{k\,(x_1 +x_2)^2}{2}=\frac{m\,v^2}{2}+\frac{k\,x_1^2}{2}+mgh_2-mgh_3}$
$\mathbf\small{\Longrightarrow \frac{k\,(x_1 +x_2)^2}{2}=\frac{m\,v^2}{2}+\frac{k\,x_1^2}{2}+mg(h_2-h_3)}$
$\mathbf\small{\Longrightarrow \frac{k\,(x_1 +x_2)^2}{2}=\frac{m\,v^2}{2}+\frac{k\,x_1^2}{2}+mgx_2}$
• Now we multiply both sides by 2. We get:
$\mathbf\small{{k\,(x_1 +x_2)^2}={m\,v^2}+{k\,x_1^2}+2mgx_2}$
• Expanding the left side, we get:
$\mathbf\small{{kx_1^2 + 2kx_1 x_2+kx_2^2}={m\,v^2}+{k\,x_1^2}+2mgx_2}$
• The first term on left side is same as the second term on the right side. So they will cancel each other. We get:
$\mathbf\small{{2kx_1 x_2+kx_2^2}={m\,v^2}+2mgx_2}$
• The first term on the left side is $\mathbf\small{{2kx_1 x_2}}$
• But from (6), $\mathbf\small{kx_1=mg}$
• Thus we get: $\mathbf\small{{2mg x_2+kx_2^2}={m\,v^2}+2mgx_2}$
• Now, the first term on left side is same as the last term on the right side
16. Thus we get: $\mathbf\small{{k\,x_2^2}={m\,v^2}}$
17. From this we get: $\mathbf\small{v=\left [\sqrt{\frac{k}{m}} \right ]x_2}$
• This is the answer to our query in step (11)
■ Note: In the above discussion, h1, h2 and h3 are unknowns. But they do not create any problem while solving. This is because, it is the 'differences between the heights' that matters. We can see that, those differences are known values: x1 and x2. Thus we can say: Any convenient level can be chosen as the datum (ground level) for defining gravitational potential energy.
By applying the Law of conservation of energy, we can solve many problems related to vertical springs. We will see a few solved examples:
Solved example 6.18
A block of mass 0.35 kg is attached to a vertical spring. The spring constant of the spring is 30 Nm-1.
The block is initially supported by hand so that the spring is neither stretched nor compressed. It is then released. How far does the ball fall? [g = 9.8 ms-2]
Solution:
1. Fig.6.39(a) below shows the situation when the block is supported by hand
 |
| Fig.6.39 |
• It is at a height 'h1' from the ground level
• If there is no support from the hand, the block will be at equilibrium at a 'stretched position' further below
2. When the block is released from the equilibrium level in fig.a, it will fall down to a lower level.
• This level is marked as ‘stretched level’ in fig.b
• It is at a height 'h2' from the ground level
• It is at a vertical distance of ‘x’ from the equilibrium
3. The block will remain at the stretched level only momentarily. Because, it will be immediately pulled up by the spring
• We are asked to find the maximum distance traveled downwards from Equilibrium level
• That is., we are asked to find 'x'
4. For that, we will use the Law of conservation of energy
• We will choose two levels:
(i) The Equilibrium
(ii) The stretched-level
■ The total energy must be the same at those two levels
5. Let us write the various energies:
(i) Total energy Ei at the Equilibrium level
• Kinetic energy = 0
• Spring potential energy = 0
♦ ∵ The spring is neither stretched nor compressed
• Gravitational potential energy = $\mathbf\small{mgh_1}$
■ So total energy Ei = $\mathbf\small{mgh_1}$
(ii) Total energy Ef at the stretched-level:
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{k\,x^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_2}$
■ So total energy Ef = $\mathbf\small{\frac{k\,x^2}{2}+mgh_2}$
6. Equating the two energies we get:
$\mathbf\small{mgh_1=\frac{k\,x^2}{2}+mgh_2}$
$\mathbf\small{\Longrightarrow mg(h_1-h_2)=\frac{k\,x^2}{2}}$
$\mathbf\small{\Longrightarrow mgx=\frac{k\,x^2}{2}}$ [∵ (h1-h2) = x]
$\mathbf\small{\Longrightarrow mg=\frac{k\,x}{2}}$
$\mathbf\small{\Longrightarrow x=\frac{2mg}{k}}$
Substituting the known values, we get: x = 0.229 m
■ Note: In this example, h1 and h2 are unknowns. But they do not create any problem while solving. This is because, it is the 'difference between h1 and h2' that matters. We can see that, this difference is 'x'. Thus we can say: Any convenient level can be chosen as the datum (ground level) for defining gravitational potential energy.
Solved example 6.19
A block of mass 5 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below so that it acquires an upward speed of 2 m/s. How high will it rise? [g = 10 ms-2]
Solution:
1. In fig.6.40 below, the natural length of the spring is indicated by 'Equilibrium-1'
 |
| Fig.6.40 |
2. Using the relation kx = mg, we get:
$\mathbf\small{k=\frac{mg}{x}=\frac{5 \times 10}{0.10}=500\,Nm^{-1}}$
3. For this problem, we need not bother about 'how' or 'why' the impulsive force was given
• All that matters is this:
♦ The block started it's upward journey from Equilibrium-2 with a velocity of 2 ms-1
♦ In other words, the initial velocity of the upward journey is 2 ms-1
4. As a result of this journey, let the block reach the level indicated as 'compressed-level' in fig.c
• Let this 'compressed-level' be at a vertical distance of 'x1' from the Equilibrium level
5. We are asked to find the total distance '(0.1+x1)'
• We can find this total distance once we find x1
6. To find x1, we use the law of conservation of energy
• We will choose two levels:
(i) The Equilibrium-2
(ii) The compressed-level
■ The total energy must be the same at those two levels
7. Let us write the various energies:
(i) Total energy Ei at the Equilibrium-2
• Kinetic energy = $\mathbf\small{\frac{mv^2}{2}=\frac{5 \times 2^2}{2}=10\,J}$
• Spring potential energy = $\mathbf\small{\frac{kx^2}{2}=\frac{500 \times 0.1^2}{2}=2.5\,J}$
• Gravitational potential energy = $\mathbf\small{mgh_2=5 \times 10 \times h_2=50h_2\,\,J}$
■ So total energy Ei = (12.5+50h2) J
(ii) Total energy Ef at the compressed-level:
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{k\,x_1^2}{2}=\frac{500\,x_1^2}{2}=250x_1^2}\,\, J$
• Gravitational potential energy = $\mathbf\small{mgh_3=5 \times 10 \times h_3=50h_3\,\,J}$
■ So total energy Ef = $\mathbf\small{250x_1^2+50h_3}$
8. Equating the two energies we get:
$\mathbf\small{12.5+50h_2=250x_1^2+50h_3}$
$\mathbf\small{\Longrightarrow 12.5=250x_1^2+50(h_3-h_2)}$
$\mathbf\small{\Longrightarrow 12.5=250x_1^2+50(x_1+0.1)}$
• Dividing both sides by 12.5, we get: $\mathbf\small{1=20x_1^2+4(x_1+0.1)}$
$\mathbf\small{\Longrightarrow 1=20x_1^2+4x_1+0.4}$
$\mathbf\small{\Longrightarrow 20x_1^2+4x_1-0.6=0}$
• This is a quadratic equation in x1
• Solving it, we get: x1 = 0.1 or -0.3
• x1 cannot be negative. So we can write: x1 = 0.1 m
■ Thus the answer is: The block rises to a height of 0.2 m above Equilibrium-2
■ Note: In this example, h1, h2 and h3 are unknowns. But they do not create any problem while solving. This is because, it is the 'difference between heights' that matters. Those 'differences' can be easily obtained from the fig. Thus we can say: Any convenient level can be chosen as the datum (ground level) for defining gravitational potential energy.
Solved example 6.20
A vertical spring with constant 200 N/m has a light platform on its top. When a 500 g mass is kept on the platform spring compresses 2.5 cm. Mass is now pushed down 7.50 cm further and released. How far above later position will the mass fly? [g = 10 ms-2]
Solution:
1. In fig.6.41 below, the natural length of the spring is indicated by 'Equilibrium-1'
 |
| Fig.6.41 |
2. Using the relation kx = mg, we get:
$\mathbf\small{k=\frac{mg}{x}=\frac{0.5 \times 10}{0.025}=200\,Nm^{-1}}$
3. Now, the block is pushed further down by 0.075 m
• This is shown in fig.c
• The new level is indicated as 'Compressed Level'
4. From this compressed position, the block is released
• The spring will then release it's potential energy and expand
• But it can go only upto the Equilibrium-1 level
• This is shown in fig.d
• The block will have acquired kinetic energy, which will enable it to fly further up above Equilibrium-1
5. The maximum height that the block achieve is shown by a white horizontal dashed line
• We want to find the vertical distance between 'Compressed Level' and 'Maximum Height'
• This vertical distance is marked as x2 in the fig.d
6. To find x2, we use the law of conservation of energy
• We will choose two levels:
(i) The compressed-level
(ii) The Maximum height
■ The total energy must be the same at those two levels
7. Let us write the various energies:
(i) Total energy Ei at the Compressed-level
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{kx^2}{2}=\frac{200 \times 0.1^2}{2}=1.0\,J}$
• Gravitational potential energy = $\mathbf\small{mgh_3=0.5 \times 10 \times h_3=5h_3\,\,J}$
■ So total energy Ei = (1+5h3) J
(ii) Total energy Ef at theMaximum height
• Kinetic energy = 0
• Spring potential energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_4=0.5 \times 10 \times h_4=5h_4\,\,J}$
■ So total energy Ef = $\mathbf\small{5h_4}$
8. Equating the two energies we get:
$\mathbf\small{1+5h_3=5h_4}$
$\mathbf\small{\Longrightarrow 1=5(h_4-h_3)=5x_2}$
$\mathbf\small{\Longrightarrow x_2=\frac{1}{5}=0.2\,m}$
So the block travels a distance of 20 cm from the point of release
Another method:
1. After the block is released, consider the travel from 'Compressed-Level' to 'Equilibrium-1'
■ The total energy must be the same at those two levels
2. Let us write the various energies:
(i) Total energy Ei at the Compressed-level
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{kx^2}{2}=\frac{200 \times 0.1^2}{2}=1.0\,J}$
• Gravitational potential energy = $\mathbf\small{mgh_3=0.5 \times 10 \times h_3=5h_3\,\,J}$
■ So total energy Ei = (1+5h3) J
(ii) Total energy Ef at the Equilibrium-1
• Kinetic energy = $\mathbf\small{\frac{mv^2}{2}=\frac{0.5 \times v^2}{2}=0.25v^2\,J}$
♦ Where 'v' is the velocity acquired by the block when it reaches Equilibrium-1
• Spring potential energy = 0
♦ ∵ The spring is neither stretched nor compressed at Equilibrium-1
• Gravitational potential energy = $\mathbf\small{mgh_1=0.5 \times 10 \times h_1=5h_1\,\,J}$
■ So total energy Ef = $\mathbf\small{0.25v^2+5h_1}$
3. Equating the two energies we get:
$\mathbf\small{1+5h_3=0.25v^2+5h_1}$
$\mathbf\small{\Longrightarrow 1+5(h_3-h_1)=0.25v^2}$
$\mathbf\small{\Longrightarrow 1+5[-(x+x_1)]=0.25v^2}$
$\mathbf\small{\Longrightarrow 1+5[-0.1]=0.25v^2}$
$\mathbf\small{\Longrightarrow 0.5=0.25v^2}$
$\mathbf\small{\Longrightarrow v=\sqrt{2}\,\,ms^{-1}}$
4. Now we have a journey from 'Equilibrium-1' to 'Max. Height'
For this journey:
• The initial velocity is $\mathbf\small{\sqrt{2}\,\,ms^{-1}}$
• Final velocity is zero
• Acceleration = -g = -10 ms-2
5. We can use the third equation of motion:
$\mathbf\small{v^2-u^2=2as}$
• Substituting the known values, we get:
$\mathbf\small{0^2-(\sqrt{2})^2=2 \times -10 \times s}$
$\mathbf\small{\Longrightarrow -2=-20s}$
$\mathbf\small{\Longrightarrow s=0.1\,\,m}$
6. Thus, the distance between ‘Equilibrium-1’ and ‘Maximum height’ is 0.1 m
■ So the required distance between ‘Compressed-level’ and ‘Maximum height’
= 0.1 + (0.025 + 0.075) = 0.2 m (same as above)
■ Note: In this example, h1, h2, h3 and h4 are unknowns. But they do not create any problem while solving. This is because, it is the 'difference between heights' that matters. Those 'differences' can be easily obtained from the fig. Thus we can say: Any convenient level can be chosen as the datum (ground level) for defining gravitational potential energy.
Solved example 6.21
A block of mass 20 kg is released from rest so as to slide in between vertical rails. It compresses a spring (k = 1920 N/m) placed 1 m below the starting point of the block. The rails offer a frictional force of 40 N which opposes the fall of the block. Find (I) The velocity of the block just before striking the spring (ii) The maximum compression of the spring (iii) The distance through which the block is rebounded up, from the maximum compressed position. [g = 10 ms-2]
Solution:
1. In fig.6.42(a) below, the block is at the ‘initial-height’
 |
| Fig.6.42 |
• The top of the spring is marked as ‘Equilibrium-1’
• We want the velocity of the block just before striking the spring
2. To find that velocity, we use the law of conservation of energy
• We will choose two levels:
(i) The Initial-Height
(ii) The Equilibrium-1
3. Let us write the various energies:
(i) Total energy Ei at the Initial-Height
• Kinetic energy = 0
• Spring potential energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_1=20 \times 10 \times h_1=200h_1\,\,J}$
■ So total energy Ei = (200h1) J
(ii) Total energy Ef at the Equilibrium-1
• Kinetic energy = $\mathbf\small{\frac{mv^2}{2}=\frac{20 \times v^2}{2}=10v^2\,J}$
♦ Where 'v' is the velocity of the block just before striking the spring
• Spring potential energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_0=20 \times 10 \times h_0=200h_0\,\,J}$
■ So total energy Ef = $\mathbf\small{10v^2+200h_0}$
4. By the law of conservation of energy, all the energy at 'Initial-height' must be available at 'Equilibrium-1'
• But some energy is lost for doing work against friction
• Quantity of that 'lost energy' = Frictional force × distance = 40 × 1 = 40 J
5. If we add this 'lost energy' on the right side, then the energies will balance. Thus we get:
$\mathbf\small{200h_1=10v^2+200h_0+40}$
$\mathbf\small{\Longrightarrow 200(h_1-h_0)-40=10v^2}$
$\mathbf\small{\Longrightarrow 200 \times 1-40=10v^2}$
$\mathbf\small{\Longrightarrow 160=10v^2}$
$\mathbf\small{\Longrightarrow v=4\,\,ms^{-1}}$
This is the answer for part (i)
6. The lowest level reached by the block is marked as ‘Compressed-Level’
• We want the vertical height between ‘Equilibrium-1’ and ‘Compressed-level’
7. To find that height, we use the law of conservation of energy
• We will choose two levels:
(i) The Equilibrium-1
(ii) The Compressed-Level
8. Let us write the various energies:
(i) Total energy Ei at the Equilibrium-1
• Kinetic energy = $\mathbf\small{\frac{mv^2}{2}=\frac{20 \times 4^2}{2}=160\,J}$
• Spring potential energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_1=20 \times 10 \times h_1=200h_0\,\,J}$
■ So total energy Ei = (160+200h0) J
(ii) Total energy Ef at the Compressed-Level
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{kx^2}{2}=\frac{1920 \times (h_0-h_2)^2}{2}=960(h_0-h_2)\,J}$
♦ For convenience, let us put (h0-h2) = x1
♦ Then Spring potential energy = 960x12
• Gravitational potential energy = $\mathbf\small{mgh_2=20 \times 10 \times h_2=200h_2\,\,J}$
■ So total energy Ef = 960x12 + 200h2
9. By the law of conservation of energy, all the energy at 'Equilibrium-1' must be available at 'Compressed-Level'
• But some energy is lost for doing work against friction
• Quantity of that 'lost energy' = Frictional force × distance = 40 × x1 J
10. If we add this 'lost energy' on the right side, then the energies will balance. Thus we get:
160+200h0 = 960x12 + 200h2 + 40x1
⇒ 200(h0-h2) = 960x12 + 40x1 - 160
⇒ 200x1 = 960x12 + 40x1 - 160
⇒ 960x12 -160x1 - 160 = 0
• Dividing both sides by 160, we get:
6x12 -x1 - 1 = 0
• This is a quadratic equation in x1. Solving it, we get:
x1 = 0.5 m or -0.333 m
• x1 Cannot be negative. So we can write x1 = 0.5 m
• Thus we get: The maximum compression of the spring = 0.5 m
• This is the answer for part (ii)
11. After reaching the 'compressed-level', the spring will expand
• The block will be pushed upwards and it will fly upto a certain height after leaving the upper end of the spring
• We want the maximum height reached by the block
• This maximum height is marked with white horizontal dashed line
• We want the vertical distance between 'compressed-level' and 'maximum-height'
12. For that, we use the law of conservation of energy
• We will choose two levels:
(i) The compressed-level
(ii) The Maximum height
■ The total energy must be the same at those two levels
13. Let us write the various energies:
(i) Total energy Ei at the Compressed-level
• Kinetic energy = 0
• Spring potential energy = $\mathbf\small{\frac{kx^2}{2}=\frac{1920 \times 0.5^2}{2}=240\,J}$
• Gravitational potential energy = $\mathbf\small{mgh_3=20 \times 10 \times h_3=200h_2\,\,J}$
■ So total energy Ei = (240+200h2) J
(ii) Total energy Ef at theMaximum height
• Kinetic energy = 0
• Spring potential energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_3=20 \times 10 \times h_3=200h_3\,\,J}$
■ So total energy Ef = $\mathbf\small{200h_3}$
14. By the law of conservation of energy, all the energy at 'compressed-level' must be available at 'maximum-height'
• But some energy is lost for doing work against friction
• Quantity of that 'lost energy' = Frictional force × distance = 40 × (h3-h2)
• For convenience, let us put (h3-h2) = x2
• So 'lost energy' = 40x2
15. If we add this 'lost energy' on the right side, then the energies will balance. Thus we get:
240+200h2 = 200h3 + 40x2
⇒ 240 = 200(h3-h2) + 40x2
⇒ 240 = 200x2 + 40x2
⇒ 240 = 240x2.
⇒ x2 = 1 m
• This is the answer for part (iii)
■ Note: In this example, h0, h1, h2 and h3 are unknowns. But they do not create any problem while solving. This is because, it is the 'difference between heights' that matters. Those 'differences' can be easily obtained from the fig. Thus we can say: Any convenient level can be chosen as the datum (ground level) for defining gravitational potential energy.
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