In the previous section, we saw the conservation of energy in the 'horizontal oscillation of a spring'. We wrote upto the 19th step. We saw that, if there is no friction or air resistance, the oscillation will continue indefinitely. We will now see the details about such an 'ideal' situation. We will start from the 20th step.
20. Assume that in an experiment, the block (in fig.6.33 that we saw in the previous section) is pulled in such a way that xt = 'd'. That is., OA = d
• For convenience, the fig.6.33 is shown again below:
We can write the following details:
(i) At A:
♦ Potential energy P = $\mathbf\small{\frac{k\,d^2}{2}}$
♦ Kinetic energy K = 0
■ Total mechanical energy at A = (P + K) = $\mathbf\small{\frac{k\,d^2}{2}}$
(ii) At O:
♦ P = 0
♦ K = $\mathbf\small{\frac{m\,v_O^2}{2}}$
♦ But using the result in 8(iii), we have:
♦ $\mathbf\small{\frac{m\,v_O^2}{2}=\frac{k\,d^2}{2}}$
♦ Thus K = $\mathbf\small{\frac{k\,d^2}{2}}$
■ Total mechanical energy at O = (P + K) = $\mathbf\small{\frac{k\,d^2}{2}}$
(The same total value at A)
(iii) At B:
♦ P = $\mathbf\small{\frac{k\,d^2}{2}}$
[The same value at A because, in the absence of resistive forces, the block will go in the same distance d. This we saw in (17)]
♦ K = 0
■ Total mechanical energy at B = (P + K) = $\mathbf\small{\frac{k\,d^2}{2}}$
(The same total value at A)
■ Thus there is indeed conservation of energy
21. We saw the conservation of energy at the extreme points (A and B) and also at the equilibrium position O
• But energy will be conserved at intermediate points also
(i) At intermediate points there will be both P and K
(ii) Let vx be the velocity of the block at any distance 'x' from O
♦ Then kinetic energy at that point (Kx) = $\mathbf\small{\frac{m\,v_x^2}{2}}$
(iii) The potential energy (Px) at that point will be $\mathbf\small{\frac{k\,x^2}{2}}$
(iv) So total mechanical energy (E) at any point which is at a distance of 'x' from 'O' will be given by:
E = (Px+Kx) = $\mathbf\small{\frac{k\,x^2}{2}+\frac{m\,v_x^2}{2}}$
(v) But we already know the total energy. It is $\mathbf\small{\frac{k\,d^2}{2}}$
♦ Where 'd' is the initial 'pulled distance'
■ Thus we have an equation:
$\mathbf\small{\frac{k\,d^2}{2}=\frac{k\,x^2}{2}+\frac{m\,v_x^2}{2}}$
• This equation gives us the total energy E at any point we select
22. With the help of the equation, we can plot the graph showing the following 3 items:
(i) Variation of P when the 'distance from O' changes
(ii) Variation of K when the 'distance from O' changes
(iii) The 'unchanging nature' of total energy E
23. It is easier to understand the procedure of this plotting, if we consider an actual example:
• The spring constant k of a spring is 5000 Nm-1.
• A block of mass 'm' is attached to the free end. It rests on a smooth horizontal surface
• This block is pulled by a distance 'd' of 8 cm and then released
■ Plot the variation of P and K
■ Also show that E [= (P+K)] is a constant
[Assume that there is no friction and air resistance]
Solution:
(i) The table shown below gives the values of P and K:
• The block is pulled by 0.08 m. Since there is no friction or air resistance, it should go in upto -.08 m on the other side of 'O'
• For convenience of plotting, the total distance of 0.16 m is divided into equal parts. Each part is 0.01 m in length
(ii) A sample calculation should clarify any doubts about how the table was prepared:
• Let the distance from 'O' be 0.05 m on the left side. So x = -0.05 m
We have:
• $\mathbf\small{E=\frac{k\,d^2}{2}=\frac{5000 \times .08^2}{2}=16}$
• $\mathbf\small{P_{(-.05)}=\frac{k\,x^2}{2}=\frac{5000 \times (-.05)^2}{2}=6.25}$
• So K(-.05) = (16-6.25) = 9.75
■ Reader may check the other points by this method
(iii) Once the table is prepared, we can plot the graphs. It is shown in fig.6.34 below:
• The yellow curve is the graph of P
♦ It is parabolic in shape
♦ The equation $\mathbf\small{P=\frac{k\,x^2}{2}}$ is a second degree equation
♦ It is indeed the equation of a parabola
• The cyan curve is the graph of K
♦ It is parabolic in shape
♦ The equation $\mathbf\small{K=\frac{k\,d^2}{2}-\frac{k\,x^2}{2}}$ is a second degree equation
♦ It is indeed the equation of a parabola
• The pink horizontal line is the graph of E
♦ It is a horizontal line through '16'
♦ Where ever we consider a point between -0.08 and +0.08, the total energy at that point will be 16 joules
23. Let us see an application of the above graph:
(i) Consider fig.6.35 below.
• A magenta vertical line is drawn at any random point say x = -0.04 m
♦ This line extends upwards until it meets the P-curve
♦ The height of this magenta line gives the potential energy at x = -0.04 m
♦ From the fig., we have: height = 4
• A white vertical line is drawn at that same point x = -0.04 m
♦ This line extends upwards until it meets the K-curve
♦ The height of this white line gives the kinetic energy at x = -0.04 m
♦ From the fig., we have: height = 12
• The total of the two heights gives the total energy E at x = -0.04 m
♦ In this case, it is: E = (4+12) = 16 joules
(ii) Another example:
• A magenta vertical line is drawn at any random point say x = 0.06 m
♦ This line extends upwards until it meets the P-curve
♦ The height of this magenta line gives the potential energy at x = 0.06 m
♦ From the fig., we have: height = 9
• A white vertical line is drawn at that same point x = 0.06 m
♦ This line extends upwards until it meets the K-curve
♦ The height of this white line gives the kinetic energy at x = 0.06 m
♦ From the fig., we have: height = 7
• The total of the two heights gives the total energy E at x = 0.06 m
♦ In this case, it is: E = (9+7) = 16 joules
24. The following two points may be noted:
(i) Every point on the P-curve lies below the horizontal through E
(ii) Every point on the K-curve lies below the horizontal through E
25. The P-curve and K-curve are complimentary
• This is because, when one increases, the other decreases
26. Finally, let us find the maximum velocity possible for the block
(i) Consider again the equation $\mathbf\small{\frac{k\,d^2}{2}=\frac{k\,x^2}{2}+\frac{m\,v_x^2}{2}}$
(ii) Two items contribute towards the total energy $\mathbf\small{\frac{k\,d^2}{2}}$
They are:
$\mathbf\small{\frac{k\,x^2}{2}}$ and $\mathbf\small{\frac{m\,v_x^2}{2}}$
(iii) They need not be making equal contributions
• If at a point, one is making a large contribution, the other will be making a small contribution
• This is because, E must remain a constant
(However, at the point where the two curves intersect, the contributions will be the same)
(iv) Consider any point along the x axis such that:
• One of those two items becomes zero
• Then at that point, the other item will be having it's own maximum possible value
• This is because the total 'E' is a constant
(v) For example:
• At the origin 'O', where x = 0, the potential energy (P) is zero
• So at that point, kinetic energy (K) will be having it's maximum possible value
• But the 'm' in the kinetic energy ($\mathbf\small{\frac{m\,v_x^2}{2}}$) is a constant
• That means at O, the velocity will be maximum
• We can write: vO = vmax.
• Thus we get: $\mathbf\small{\frac{k\,d^2}{2}=0+\frac{m\,v_O^2}{2}}$
$\mathbf\small{\Longrightarrow \frac{k\,d^2}{2}=\frac{m\,v_{max}^2}{2}}$
$\mathbf\small{\Longrightarrow {k\,d^2}={m\,v_{max}^2}}$
$\mathbf\small{\Longrightarrow {v_{max}^2=\frac{k}{m}\,d^2}}$
$\mathbf\small{\Longrightarrow {v_{max}=\left[\sqrt{\frac{k}{m}}\right]\,d}}$
■ Thus we can write:
• If the block is pulled by a distance 'd' from the equilibrium point 'O' and then released,
♦ At O, the block will attain the maximum possible velocity
♦ This velocity is equal to $\mathbf\small{{\left[\sqrt{\frac{k}{m}}\right]\,d}}$
(vi) Let us do a dimensional analysis of $\mathbf\small{{\left[\sqrt{\frac{k}{m}}\right]\,d}}$:
• 'k' has the unit Nm-1
• So the dimensions of 'k' is: [MLT-2] /[L] = [MT-2]
• So dimensions of $\mathbf\small{\frac{k}{m}}$ is [T-2]
• So dimensions of $\mathbf\small{\sqrt{\frac{k}{m}}}$ is [T-1]
• So dimensions of $\mathbf\small{\sqrt{\frac{k}{m}}\,d}$ is [LT-1]
• This is the dimensions of velocity
■ Thus the equation $\mathbf\small{{v_{max}=\left[\sqrt{\frac{k}{m}}\right]\,d}}$ is dimensionally correct
Now we will see some solved examples:
Solved example 6.16:
A block of mass of 1.4 kg rests on a horizontal surface. It is pushed against a horizontal spring with k value 140 Nm-1. The other end of the spring is attached to a vertical wall. The spring gets compressed by 22 cm from it's equilibrium position (See fig.6.36.a below). When the block is released from this compressed position, it travels a distance of 105 cm before coming to rest.
Calculate the coefficient of kinetic friction between the block and the horizontal surface. Neglect air resistance. Take g = 9.8 ms-2
Solution:
1. Given that air resistance can be neglected. Let us consider the situation where there is no friction also
• When the spring is compressed, potential energy is stored in it (There is no kinetic energy in the compressed position)
• When the spring is released, this potential energy is converted into kinetic energy of the block
• If there is no friction, the block will continue to move indefinitely with a velocity 'v'
2. But since there is friction, the kinetic energy acquired will be used up to do work against friction
• It is clear that, all the kinetic energy is used up when it travels 105 cm from the point of release
3. Work done against friction = Frictional force × distance
= μkmg × 1.05 = μk×1.4×9.8×1.05 = 14.406 μk joules
4. So the acquired kinetic energy is 14.406 μk joules
Then the initial potential energy will also be equal to 14.406 μk joules
5. But the initial potential energy = $\mathbf\small{\frac{k\,x^2}{2}=\frac{140 \times 0.22^2}{2}=3.388\,joules}$
So we can write: 14.406 μk = 3.388
Thus μk = 0.235179786
Another method:
• The method that we saw above used energies only.
• The method that we will see next, will help us to demonstrate that, the same result can be obtained by using forces also
1. When the block is released, the spring will push it towards the right
• The block travels for a distance of 1.05 m
• But the spring cannot go that much distance
• The end of the spring can go only up to the equilibrium position
• That means, the 'physical contact between the spring and the block' will be present only up to 22 cm
• After that, the block is on it's own. It gets no further supply of energy
• This is shown in fig.6.36(b)
2. We can write two points:
(i) The block acquires a 'certain amount of energy' at the end of the first 22 cm journey
(ii) This acquired energy will be used up (for doing work against friction) in the next 83 cm journey
3. How much energy does the block acquire at the end of the first 22 cm journey?
Let us find out:
• If there is no friction, this energy must be equal to 'the initial potential energy stored in the spring'
• But the initial potential energy = $\mathbf\small{\frac{k\,x^2}{2}=\frac{140 \times 0.22^2}{2}=3.388\,joules}$
20. Assume that in an experiment, the block (in fig.6.33 that we saw in the previous section) is pulled in such a way that xt = 'd'. That is., OA = d
• For convenience, the fig.6.33 is shown again below:
 |
| Fig.6.33 |
(i) At A:
♦ Potential energy P = $\mathbf\small{\frac{k\,d^2}{2}}$
♦ Kinetic energy K = 0
■ Total mechanical energy at A = (P + K) = $\mathbf\small{\frac{k\,d^2}{2}}$
(ii) At O:
♦ P = 0
♦ K = $\mathbf\small{\frac{m\,v_O^2}{2}}$
♦ But using the result in 8(iii), we have:
♦ $\mathbf\small{\frac{m\,v_O^2}{2}=\frac{k\,d^2}{2}}$
♦ Thus K = $\mathbf\small{\frac{k\,d^2}{2}}$
■ Total mechanical energy at O = (P + K) = $\mathbf\small{\frac{k\,d^2}{2}}$
(The same total value at A)
(iii) At B:
♦ P = $\mathbf\small{\frac{k\,d^2}{2}}$
[The same value at A because, in the absence of resistive forces, the block will go in the same distance d. This we saw in (17)]
♦ K = 0
■ Total mechanical energy at B = (P + K) = $\mathbf\small{\frac{k\,d^2}{2}}$
(The same total value at A)
■ Thus there is indeed conservation of energy
21. We saw the conservation of energy at the extreme points (A and B) and also at the equilibrium position O
• But energy will be conserved at intermediate points also
(i) At intermediate points there will be both P and K
(ii) Let vx be the velocity of the block at any distance 'x' from O
♦ Then kinetic energy at that point (Kx) = $\mathbf\small{\frac{m\,v_x^2}{2}}$
(iii) The potential energy (Px) at that point will be $\mathbf\small{\frac{k\,x^2}{2}}$
(iv) So total mechanical energy (E) at any point which is at a distance of 'x' from 'O' will be given by:
E = (Px+Kx) = $\mathbf\small{\frac{k\,x^2}{2}+\frac{m\,v_x^2}{2}}$
(v) But we already know the total energy. It is $\mathbf\small{\frac{k\,d^2}{2}}$
♦ Where 'd' is the initial 'pulled distance'
■ Thus we have an equation:
$\mathbf\small{\frac{k\,d^2}{2}=\frac{k\,x^2}{2}+\frac{m\,v_x^2}{2}}$
• This equation gives us the total energy E at any point we select
22. With the help of the equation, we can plot the graph showing the following 3 items:
(i) Variation of P when the 'distance from O' changes
(ii) Variation of K when the 'distance from O' changes
(iii) The 'unchanging nature' of total energy E
23. It is easier to understand the procedure of this plotting, if we consider an actual example:
• The spring constant k of a spring is 5000 Nm-1.
• A block of mass 'm' is attached to the free end. It rests on a smooth horizontal surface
• This block is pulled by a distance 'd' of 8 cm and then released
■ Plot the variation of P and K
■ Also show that E [= (P+K)] is a constant
[Assume that there is no friction and air resistance]
Solution:
(i) The table shown below gives the values of P and K:
• The block is pulled by 0.08 m. Since there is no friction or air resistance, it should go in upto -.08 m on the other side of 'O'
• For convenience of plotting, the total distance of 0.16 m is divided into equal parts. Each part is 0.01 m in length
• Let the distance from 'O' be 0.05 m on the left side. So x = -0.05 m
We have:
• $\mathbf\small{E=\frac{k\,d^2}{2}=\frac{5000 \times .08^2}{2}=16}$
• $\mathbf\small{P_{(-.05)}=\frac{k\,x^2}{2}=\frac{5000 \times (-.05)^2}{2}=6.25}$
• So K(-.05) = (16-6.25) = 9.75
■ Reader may check the other points by this method
(iii) Once the table is prepared, we can plot the graphs. It is shown in fig.6.34 below:
 |
| Fig.6.34 |
♦ It is parabolic in shape
♦ The equation $\mathbf\small{P=\frac{k\,x^2}{2}}$ is a second degree equation
♦ It is indeed the equation of a parabola
• The cyan curve is the graph of K
♦ It is parabolic in shape
♦ The equation $\mathbf\small{K=\frac{k\,d^2}{2}-\frac{k\,x^2}{2}}$ is a second degree equation
♦ It is indeed the equation of a parabola
• The pink horizontal line is the graph of E
♦ It is a horizontal line through '16'
♦ Where ever we consider a point between -0.08 and +0.08, the total energy at that point will be 16 joules
23. Let us see an application of the above graph:
(i) Consider fig.6.35 below.
• A magenta vertical line is drawn at any random point say x = -0.04 m
♦ This line extends upwards until it meets the P-curve
♦ The height of this magenta line gives the potential energy at x = -0.04 m
♦ From the fig., we have: height = 4
 |
| Fig.6.35 |
♦ This line extends upwards until it meets the K-curve
♦ The height of this white line gives the kinetic energy at x = -0.04 m
♦ From the fig., we have: height = 12
• The total of the two heights gives the total energy E at x = -0.04 m
♦ In this case, it is: E = (4+12) = 16 joules
(ii) Another example:
• A magenta vertical line is drawn at any random point say x = 0.06 m
♦ This line extends upwards until it meets the P-curve
♦ The height of this magenta line gives the potential energy at x = 0.06 m
♦ From the fig., we have: height = 9
• A white vertical line is drawn at that same point x = 0.06 m
♦ This line extends upwards until it meets the K-curve
♦ The height of this white line gives the kinetic energy at x = 0.06 m
♦ From the fig., we have: height = 7
• The total of the two heights gives the total energy E at x = 0.06 m
♦ In this case, it is: E = (9+7) = 16 joules
24. The following two points may be noted:
(i) Every point on the P-curve lies below the horizontal through E
(ii) Every point on the K-curve lies below the horizontal through E
25. The P-curve and K-curve are complimentary
• This is because, when one increases, the other decreases
26. Finally, let us find the maximum velocity possible for the block
(i) Consider again the equation $\mathbf\small{\frac{k\,d^2}{2}=\frac{k\,x^2}{2}+\frac{m\,v_x^2}{2}}$
(ii) Two items contribute towards the total energy $\mathbf\small{\frac{k\,d^2}{2}}$
They are:
$\mathbf\small{\frac{k\,x^2}{2}}$ and $\mathbf\small{\frac{m\,v_x^2}{2}}$
(iii) They need not be making equal contributions
• If at a point, one is making a large contribution, the other will be making a small contribution
• This is because, E must remain a constant
(However, at the point where the two curves intersect, the contributions will be the same)
(iv) Consider any point along the x axis such that:
• One of those two items becomes zero
• Then at that point, the other item will be having it's own maximum possible value
• This is because the total 'E' is a constant
(v) For example:
• At the origin 'O', where x = 0, the potential energy (P) is zero
• So at that point, kinetic energy (K) will be having it's maximum possible value
• But the 'm' in the kinetic energy ($\mathbf\small{\frac{m\,v_x^2}{2}}$) is a constant
• That means at O, the velocity will be maximum
• We can write: vO = vmax.
• Thus we get: $\mathbf\small{\frac{k\,d^2}{2}=0+\frac{m\,v_O^2}{2}}$
$\mathbf\small{\Longrightarrow \frac{k\,d^2}{2}=\frac{m\,v_{max}^2}{2}}$
$\mathbf\small{\Longrightarrow {k\,d^2}={m\,v_{max}^2}}$
$\mathbf\small{\Longrightarrow {v_{max}^2=\frac{k}{m}\,d^2}}$
$\mathbf\small{\Longrightarrow {v_{max}=\left[\sqrt{\frac{k}{m}}\right]\,d}}$
■ Thus we can write:
• If the block is pulled by a distance 'd' from the equilibrium point 'O' and then released,
♦ At O, the block will attain the maximum possible velocity
♦ This velocity is equal to $\mathbf\small{{\left[\sqrt{\frac{k}{m}}\right]\,d}}$
(vi) Let us do a dimensional analysis of $\mathbf\small{{\left[\sqrt{\frac{k}{m}}\right]\,d}}$:
• 'k' has the unit Nm-1
• So the dimensions of 'k' is: [MLT-2] /[L] = [MT-2]
• So dimensions of $\mathbf\small{\frac{k}{m}}$ is [T-2]
• So dimensions of $\mathbf\small{\sqrt{\frac{k}{m}}}$ is [T-1]
• So dimensions of $\mathbf\small{\sqrt{\frac{k}{m}}\,d}$ is [LT-1]
• This is the dimensions of velocity
■ Thus the equation $\mathbf\small{{v_{max}=\left[\sqrt{\frac{k}{m}}\right]\,d}}$ is dimensionally correct
Now we will see some solved examples:
Solved example 6.16:
A block of mass of 1.4 kg rests on a horizontal surface. It is pushed against a horizontal spring with k value 140 Nm-1. The other end of the spring is attached to a vertical wall. The spring gets compressed by 22 cm from it's equilibrium position (See fig.6.36.a below). When the block is released from this compressed position, it travels a distance of 105 cm before coming to rest.
 |
| Fig.6.36 |
Solution:
1. Given that air resistance can be neglected. Let us consider the situation where there is no friction also
• When the spring is compressed, potential energy is stored in it (There is no kinetic energy in the compressed position)
• When the spring is released, this potential energy is converted into kinetic energy of the block
• If there is no friction, the block will continue to move indefinitely with a velocity 'v'
2. But since there is friction, the kinetic energy acquired will be used up to do work against friction
• It is clear that, all the kinetic energy is used up when it travels 105 cm from the point of release
3. Work done against friction = Frictional force × distance
= μkmg × 1.05 = μk×1.4×9.8×1.05 = 14.406 μk joules
4. So the acquired kinetic energy is 14.406 μk joules
Then the initial potential energy will also be equal to 14.406 μk joules
5. But the initial potential energy = $\mathbf\small{\frac{k\,x^2}{2}=\frac{140 \times 0.22^2}{2}=3.388\,joules}$
So we can write: 14.406 μk = 3.388
Thus μk = 0.235179786
Another method:
• The method that we saw above used energies only.
• The method that we will see next, will help us to demonstrate that, the same result can be obtained by using forces also
1. When the block is released, the spring will push it towards the right
• The block travels for a distance of 1.05 m
• But the spring cannot go that much distance
• The end of the spring can go only up to the equilibrium position
• That means, the 'physical contact between the spring and the block' will be present only up to 22 cm
• After that, the block is on it's own. It gets no further supply of energy
• This is shown in fig.6.36(b)
2. We can write two points:
(i) The block acquires a 'certain amount of energy' at the end of the first 22 cm journey
(ii) This acquired energy will be used up (for doing work against friction) in the next 83 cm journey
3. How much energy does the block acquire at the end of the first 22 cm journey?
Let us find out:
• If there is no friction, this energy must be equal to 'the initial potential energy stored in the spring'
• But the initial potential energy = $\mathbf\small{\frac{k\,x^2}{2}=\frac{140 \times 0.22^2}{2}=3.388\,joules}$
• That means., if there is no friction, the 'kinetic energy at the end of 22 cm' = 3.388 joules
4. But the actual kinetic energy will be less than 3.388 joules. This is because, some energy is lost for doing work against friction in the 22 cm length
• This work done against friction = Frictional force × distance
= μkmg × 0.22 = μk×1.4×9.8×0.22 = 3.0184 μk joules
5. If we add this much energy to the right side of the equation, the energies will balance
• So we get: $\mathbf\small{3.388=\frac{m\,v_O^2}{2}+3.0184\,\mu_k}$
$\mathbf\small{\Longrightarrow 3.388=\frac{1.4\,v_O^2}{2}+3.0184\,\mu_k}$
$\mathbf\small{\Longrightarrow 3.388=0.7v_O^2+3.0184\,\mu_k}$
6. Now consider the distance traveled after the 22 cm
• After the 22 cm, the block travels 83 cm
• For this 83 cm, vO is the initial velocity
• The final velocity for this 83 cm travel is zero
• The acceleration for this 83 cm travel, is the 'negative acceleration' arising due to friction
• 'Negative acceleration' arising due to friction = $\mathbf\small{\frac{\text{Frictional force}}{\text{mass}}=\frac{\mu_k\,mg}{m}=\mu_k\,g}$ = 9.8 μk.
7. We can use the equation: $\mathbf\small{v^2-u^2=2as}$
• Substituting known values, we get: $\mathbf\small{0-v_O^2=2\times (-9.8\, \mu_k)\times 0.83}$
$\mathbf\small{\Longrightarrow v_O^2=16.268 \, \mu_k}$
8. We can put this in the place of $\mathbf\small{v_O^2}$ in (5). We get:
$\mathbf\small{3.388=0.7 \times 16.268 \, \mu_k+3.0184\,\mu_k}$
$\mathbf\small{3.388=0.7 \times 16.268 \, \mu_k+3.0184\,\mu_k}$
$\mathbf\small{\Longrightarrow 3.388=14.406\,\mu_k}$
$\mathbf\small{\Longrightarrow \mu_k=0.235179786}$
• This is the same value that we obtained by the first method
• More decimal places are shown in the answers. This is to check and confirm that, both methods give the exact same answers
Solved example 6.17
A block of mass of 8 kg rests on a horizontal surface. It is pushed against a horizontal spring with k value 750 Nm-1. The other end of the spring is attached to a vertical wall. The spring gets compressed by 2.5 m from it's equilibrium position (See fig.6.37.a below).
When the block is released from this compressed position, what will be it's velocity at a distance of 25 m from the point of release? Neglect air resistance. [μk = 0.28, g = 9.8 ms-2]
Solution:
1. Given that air resistance can be neglected. Let us consider the situation where there is no friction also
• When the spring is compressed, potential energy is stored in it (There is no kinetic energy in the compressed position)
• When the spring is released, this potential energy is converted into kinetic energy of the block
• If there is no friction, the block will continue to move indefinitely with a velocity 'v'
2. But since there is friction, the kinetic energy acquired will be used up to do work against friction
• The position where the block comes to rest is not given
• Let us assume that, all the kinetic energy is used up when it travels 'y' m from the point of release
3. We can write:
• Initial potential energy = work done (against friction) while travelling 'y' m
• Work done against friction = Frictional force × distance
= μkmg × y = 0.28×8×9.8×x = 21.952 y joules
4. Thus we get: Initial potential energy = $\mathbf\small{\frac{k\,x^2}{2}=\frac{750 \times 2.5^2}{2}=21.952y\,joules}$
• So y = 106.767 m
• The block will travel 106.767 m from the point of release
• Thus it is clear that, the block will travel well beyond the 'required point' which is only 25 m from the point of release
5. We want the velocity of the block when it just passes the 'point which is 25 m away from the point of release'
• In ideal conditions, the 'initial potential energy of the spring' will be exactly equal to the 'kinetic energy of the block' at which ever point we take along it's path
• In such a condition, we can write: $\mathbf\small{\frac{k\,x^2}{2}=\frac{m\,v_{25}^2}{2}}$
♦ Where v25 is the velocity at the 25 m point
6. But the actual kinetic energy will be less than $\mathbf\small{\frac{k\,x^2}{2}}$ joules. This is because, some energy is lost for doing work against friction in the 25 m length
• More decimal places are shown in the answers. This is to check and confirm that, both methods give the exact same answers
Solved example 6.17
A block of mass of 8 kg rests on a horizontal surface. It is pushed against a horizontal spring with k value 750 Nm-1. The other end of the spring is attached to a vertical wall. The spring gets compressed by 2.5 m from it's equilibrium position (See fig.6.37.a below).
When the block is released from this compressed position, what will be it's velocity at a distance of 25 m from the point of release? Neglect air resistance. [μk = 0.28, g = 9.8 ms-2]
Solution:
1. Given that air resistance can be neglected. Let us consider the situation where there is no friction also
• When the spring is compressed, potential energy is stored in it (There is no kinetic energy in the compressed position)
• When the spring is released, this potential energy is converted into kinetic energy of the block
• If there is no friction, the block will continue to move indefinitely with a velocity 'v'
2. But since there is friction, the kinetic energy acquired will be used up to do work against friction
• The position where the block comes to rest is not given
• Let us assume that, all the kinetic energy is used up when it travels 'y' m from the point of release
3. We can write:
• Initial potential energy = work done (against friction) while travelling 'y' m
• Work done against friction = Frictional force × distance
= μkmg × y = 0.28×8×9.8×x = 21.952 y joules
4. Thus we get: Initial potential energy = $\mathbf\small{\frac{k\,x^2}{2}=\frac{750 \times 2.5^2}{2}=21.952y\,joules}$
• So y = 106.767 m
• The block will travel 106.767 m from the point of release
• Thus it is clear that, the block will travel well beyond the 'required point' which is only 25 m from the point of release
5. We want the velocity of the block when it just passes the 'point which is 25 m away from the point of release'
• In ideal conditions, the 'initial potential energy of the spring' will be exactly equal to the 'kinetic energy of the block' at which ever point we take along it's path
• In such a condition, we can write: $\mathbf\small{\frac{k\,x^2}{2}=\frac{m\,v_{25}^2}{2}}$
♦ Where v25 is the velocity at the 25 m point
6. But the actual kinetic energy will be less than $\mathbf\small{\frac{k\,x^2}{2}}$ joules. This is because, some energy is lost for doing work against friction in the 25 m length
• This work done against friction = Frictional force × distance
= μkmg × 25 = 0.28×8×9.8×25 = 548.8 joules
7. If we add this much energy to the right side of the equation, the energies will balance
So we get: $\mathbf\small{\frac{k\,x^2}{2}=\frac{m\,v_{25}^2}{2}+548.8}$
Substituting the known values, we get: $\mathbf\small{\frac{750 \times 2.5^2}{2}=\frac{8\times v_{25}^2}{2}+548.8}$
$\mathbf\small{\Longrightarrow v_{25}^2=448.7375}$
$\mathbf\small{\Longrightarrow v_{25}=21.1834251}$ ms-1.
Another method:
• The method that we saw above used energies only.
• The method that we will see next, will help us to demonstrate that, the same result can be obtained by using forces also
1. When the block is released, the spring will push it towards the right
• The block travels for a large distance. We found that this distance is of 106.767 m
• But the spring cannot go that much distance
• The end of the spring can go only up to the equilibrium position
• That means, the 'physical contact between the spring and the block' will be present only up to 2.5 m
• After that, the block is on it's own. It gets no further supply of energy
• This is shown in fig.6.37(b)
2. We can write two points:
(i) The block acquires a 'certain amount of energy' at the end of the first 2.5 m journey
(ii) This acquired energy will be used up (for doing work against friction) in the next (106.767-2.5) = 104.267 m
3. How much energy does the block acquire at the end of the first 2.5 m journey?
Let us find out:
• If there is no friction, this energy must be equal to 'the potential energy stored in the spring'
• But the initial potential energy = $\mathbf\small{\frac{k\,x^2}{2}=\frac{750 \times 2.5^2}{2}=2343.75\,joules}$
So we get: $\mathbf\small{\frac{k\,x^2}{2}=\frac{m\,v_{25}^2}{2}+548.8}$
Substituting the known values, we get: $\mathbf\small{\frac{750 \times 2.5^2}{2}=\frac{8\times v_{25}^2}{2}+548.8}$
$\mathbf\small{\Longrightarrow v_{25}^2=448.7375}$
$\mathbf\small{\Longrightarrow v_{25}=21.1834251}$ ms-1.
Another method:
• The method that we saw above used energies only.
• The method that we will see next, will help us to demonstrate that, the same result can be obtained by using forces also
1. When the block is released, the spring will push it towards the right
• The block travels for a large distance. We found that this distance is of 106.767 m
• But the spring cannot go that much distance
• The end of the spring can go only up to the equilibrium position
• That means, the 'physical contact between the spring and the block' will be present only up to 2.5 m
• After that, the block is on it's own. It gets no further supply of energy
• This is shown in fig.6.37(b)
2. We can write two points:
(i) The block acquires a 'certain amount of energy' at the end of the first 2.5 m journey
(ii) This acquired energy will be used up (for doing work against friction) in the next (106.767-2.5) = 104.267 m
3. How much energy does the block acquire at the end of the first 2.5 m journey?
Let us find out:
• If there is no friction, this energy must be equal to 'the potential energy stored in the spring'
• But the initial potential energy = $\mathbf\small{\frac{k\,x^2}{2}=\frac{750 \times 2.5^2}{2}=2343.75\,joules}$
• That means., if there is no friction, the 'kinetic energy at the end of 2.5 m' = 2343.75 joules
4. But the actual kinetic energy will be less than 2343.75 joules. This is because, some energy is lost for doing work against friction in the 2.5 m length
• This work done against friction = Frictional force × distance
= μkmg × 2.5 = 0.28×8×9.8×2.5 = 54.88 joules
5. If we add this much energy to the right side of the equation, the energies will balance
• So we get: $\mathbf\small{2343.75=\frac{8 \times v_{2.5}^2}{2}+54.88}$
$\mathbf\small{\Longrightarrow v_{2.5}^2=572.2175}$
6. Now consider the distance traveled after the 2.5 m
• More decimal places are shown in the answers. This is to check and confirm that, both methods give the exact same answers
• After the 2.5 m, the block travels (25-2.5) = 22.5 m
• For this 22.5 m, v2.5 is the initial velocity
• We want the final velocity v25 for this 22.5 m travel
• The acceleration for this 22.5 m travel, is the 'negative acceleration' arising due to friction
• 'Negative acceleration' arising due to friction = $\mathbf\small{\frac{\text{Frictional force}}{\text{mass}}=\frac{\mu_k\,mg}{m}=\mu_k\,g}$ = 9.8 × 0.28 = 2.744 ms-2.
7. We can use the equation: $\mathbf\small{v^2-u^2=2as}$
• Substituting known values, we get: $\mathbf\small{572.2175-v_{25}^2=2\times 2.744 \times 22.5}$
$\mathbf\small{\Longrightarrow v_{25}=21.1834251}$• More decimal places are shown in the answers. This is to check and confirm that, both methods give the exact same answers
No comments:
Post a Comment