In the previous section, we have completed a discussion on conservation of mechanical energy. We also saw a number of solved examples. We will see a few more in this section.
Solved example 6.22
To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed of 18 km h-1 on a smooth road and colliding with a horizontally mounted spring of spring constant 6.25 × 103 Nm-1.
1. What is the maximum compression of the spring?
2. What will be the maximum compression if friction is also taken into account. Coefficient of kinetic friction is 0.5
Solution:
1. We know that 36 kmh-1 is 10 ms-1
$\mathbf\small{\because 36 \times \left( \frac{1000}{60 \times 60} \right)=10}$
• 18 is half of 36. So 18 kmh-1 is 5 ms-1
2. Fig.6.57 below shows the simulation:
• The car moving with a certain velocity will be having kinetic energy
• After compression of the spring, the velocity of the car becomes zero
• That means, after compression, the kinetic energy is zero
3. But the initial kinetic energy is not lost.
• Based on the law of conservation of energy, the initial kinetic energy is transformed into another form of energy: The spring potential energy
4. Now we can write the equations:
• Initial kinetic energy = $\mathbf\small{\frac{mv^2}{2}=\frac{1000 \times 5^2}{2}\,\,J}$
• Let ‘x’ be the compression of the spring
• Then potential energy acquired by the spring = $\mathbf\small{\frac{kx^2}{2}=\frac{6250 \times x^2}{2}\,\,J}$
5. Equating the two energies we get: $\mathbf\small{\frac{1000 \times 25}{2}=\frac{6250 \times x^2}{2}\,\,J}$
• Thus we get x = 2 m (This is the answer for part 1)
6. When friction is also present, we cannot just equate the two energies as in (5)
• The original energy will be the same $\mathbf\small{\frac{1000 \times 5^2}{2}\,\,J}$
• But some energy will be lost for doing work against friction
• So the car will not be able to make the same compression
7. Let x1 be the new compression
• Work done against friction during the travel through this 'x1' m
= frictional force × distance = $\mathbf\small{\mu_k \times mg \times x_1=0.5 \times 1000 \times 10 \times x_1=5000x_1 \,\, J}$
8. If we add this energy to the right side, the energies will balance. So we get:
$\mathbf\small{\frac{1000 \times 5^2}{2}=\frac{6250 \times x_1^2}{2}+5000x_1}$
$\mathbf\small{\Longrightarrow 25000=6250x_1^2+10000x_1}$
$\mathbf\small{\Longrightarrow 25=6.250x_1^2+10x_1}$
$\mathbf\small{\Longrightarrow 6.250x_1^2+10x_1-25=0}$
Solving this quadratic equation, we get:
x1= 1.35 m (This is the answer for part 2)
Another approach using force method:
Part 1:
1. The car makes initial contact with the spring with a velocity of u = 5 ms-1
• The final velocity of the car v = 0
• The acceleration 'a' experienced by the car = resistive force⁄mass
2. Resistive force is the force exerted by the spring
♦ But such a force is not uniform
♦ Because it depends on the compression 'x'
3. So we take the average force:
$\mathbf\small{\frac{0+kx}{2}=0.5kx =0.5 \times 6250 \times x=3125x \,\,\text{N} }$
• So a = $\mathbf\small{\frac{3125x}{1000}=3.125x\,\,\text{ms}^{-2}}$
• Note that, this is a negative acceleration
4. We can use the equation: $\mathbf\small{v^2-u^2=2as}$
• Substituting the known values, we get: $\mathbf\small{0^2-5^2=2 \times (-3.125x) \times x}$
$\mathbf\small{\Longrightarrow -25=-6.25x^2}$
• So x = 2.0 m
• This is the same answer that we obtained before
5. Consider the following two instants:
(i) Instant at which the car makes initial contact with the spring
(ii) Instant at which the car comes to a stop
• We want the time interval between the above two instants
6. Let us use the equation: v = u + at
• Substituting the known values, we get: 0 = 5 + (-3.125x)t
⇒ 0 = 5 - 6.25t
⇒ t = 0.8 seconds
7. While using the 'energy method', we are not able to find 'time'
• We are now able to find 'time' because we applied Newton's second law and the equations of motion
Part 2:
1. The car makes initial contact with the spring with a velocity of u = 5 ms-1
• The final velocity of the car v = 0
• The acceleration 'a' experienced by the car = resistive force⁄mass
2. Resistive force = the force exerted by the spring + frictional force
• We have already calculated the force exerted by the spring as 3125x1
• Frictional force = μkmg = 0.5×1000×10 = 5000 N
• So total resistive force = (3125x1+5000) N
• So negative acceleration = Force⁄mass = $\mathbf\small{\frac{3125x_1+5000}{1000}}$
4. We can use the equation: $\mathbf\small{v^2-u^2=2as}$
• Substituting the known values, we get: $\mathbf\small{0^2-5^2=2 \times -1 \times(\frac{3125x_1+5000}{1000}) \times x_1}$
$\mathbf\small{\Longrightarrow -25000=-6250x_1^2-10000x_1}$
$\mathbf\small{\Longrightarrow 6250x_1^2+10000x_1 -25000=0}$
$\mathbf\small{\Longrightarrow 6.25x_1^2+10x_1-25=0}$
• Solving this quadratic equation, we get:
x1= 1.35 m
• This is the same answer that we obtained before
5. Consider the following two instants:
(i) Instant at which the car makes initial contact with the spring
(ii) Instant at which the car comes to a stop
• We want the time interval between the above two instants
6. Let us use the equation: v = u + at
• Substituting the known values, we get: $\mathbf\small{0=5+-1 \times (\frac{3125x_1+5000}{1000})t_1}$
$\mathbf\small{\Longrightarrow 0=5+-1 \times (\frac{3125 \times 1.35+5000}{1000})t_1}$
$\mathbf\small{\Longrightarrow 0=5+-1 \times (\frac{4218.75+5000}{1000})t_1}$
$\mathbf\small{\Longrightarrow 5000=9218.75t_1}$
So t1 = 0.5423 seconds
• Note that, when 'resistance to motion' increases, the 'time required to bring the car to a stop' decreases
7. While using the 'energy method', we are not able to find 'time'
• In 'force method', we are able to find 'time' because we apply Newton's second law and the equations of motion
Solved example 6.23
A block of mass 8 kg is released from the top of an inclined smooth surface. See fig.6.58(a) below:
The spring constant of the spring is 200 Nm-1. The block comes to rest after compressing the spring by 1 m. Find the total distance traveled by the block before it comes to rest. [g = 10 ms-2]
Solution:
1. Let 'A' be the point of release of the block
'A' is at a height of 'h1' from the datum. This is shown in fig.b
2. After compressing the spring, the block comes to rest at 'B'
'B' is at a height of 'h2' from the datum
3. By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EB
4. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_1}$
• Spring potential energy = 0
■ So total energy EA = $\mathbf\small{mgh_1}$
(ii) Total energy EB
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_2}$
• Spring potential energy = $\mathbf\small{\frac{k\,x^2}{2}}$
■ So total energy EB = $\mathbf\small{mgh_2+\frac{k\,x^2}{2}}$
5. Equating the two energies we get:
$\mathbf\small{mgh_1=mgh_2+\frac{k\,x^2}{2}}$
$\mathbf\small{\Longrightarrow mg(h_1-h_2)=\frac{k\,x^2}{2}}$
• Substituting the known values, we get: $\mathbf\small{8\times 10(h_1-h_2)=\frac{200 \times \,1^2}{2}}$
• Thus we get: (h1-h2) = 1.25 m
6. Draw a vertical through A
• Also draw a horizontal through B
• These vertical and horizontal will meet at a point. Let us call it 'C'
• A, B and C forms a right triangle as shown in fig.c
• Obviously, (h1-h2) = AC = 1.25 m
• Also, angle at B = 30o
7. In the triangle ABC, sin 30 = 1.25⁄AB
So we get: AB = 1.25⁄sin 30 = 1.25⁄0.5 = 2.5 m
Solved example 6.24
In fig.6.59(a) below, the spring constant of the spring is 1400 Nm-1. It is compressed by 0.1 m using the block and then released.
What is the total distance traveled by the block if:
1. There is no friction
2. If the coefficient of kinetic friction μk = 0.4
[g = 9.8 ms-2]
Solution:
1. Let 'A' be the point of release of the block
• 'A' is at a height of 'h1' from the datum. This is shown in fig.b
2. After release, the block comes to rest at 'B'
• 'B' is at a height of 'h2' from the datum
3. By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EB
4. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_1}$
• Spring potential energy = $\mathbf\small{\frac{k\,x^2}{2}}$
■ So total energy EA = $\mathbf\small{mgh_1+\frac{k\,x^2}{2}}$
(ii) Total energy EB
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_2}$
• Spring potential energy = 0
■ So total energy EB = $\mathbf\small{mgh_2}$
5. Equating the two energies we get:
$\mathbf\small{mgh_1+\frac{k\,x^2}{2}=mgh_2}$
$\mathbf\small{\Longrightarrow mg(h_2-h_1)=\frac{k\,x^2}{2}}$
• Substituting the known values, we get: $\mathbf\small{0.2\times 9.8(h_2-h_1)=\frac{1400 \times \,0.1^2}{2}}$
• Thus we get: (h2-h1) = 3.571 m
6. Draw a vertical through A
• Also draw a horizontal through B
• These vertical and horizontal will meet at a point. Let us call it 'C'
• A, B and C forms a right triangle as shown in fig.c
• Obviously, (h2-h1) = BC = 3.571 m
• Also, angle at A = 60o
7. In the triangle ABC, sin 60 = 3.571⁄AB
So we get: AB = 3.571⁄sin 60 = 4.124 m (This is the answer for part 1)
8. If there is no friction, EB must be equal to EA
• But since there is friction, EB will be less than EA.
• This is because, some energy is lost for doing work against friction
• If we add this 'lost energy' to EB, the energies will balance. That is:
EA = EB + Energy for doing work against friction
9. So our next task is to find the 'lost energy'
• For that, we have to first find the frictional force
• We have already learnt to find that in the case of inclined planes (Details here)
• So frictional force = μk × FN = (μk × mgcosθ) = (0.4 × 0.2 × 9.8 × cos 60) = 0.392 N
10. So energy lost = Frictional force × Distance
• This time, the block will not reach up to 'B' which we saw in the fig.6.59(b) above
• Let us call the new point 'B1' This is shown in fig.6.60(a) below:
• So the distance traveled is AB1
• Thus 'energy lost' = 0.392AB1
11. Next we write the values of EA and EB
(Note that, EB is now EB1)
(i) The position 'A' has not changed. So EA is the same $\mathbf\small{mgh_1+\frac{k\,x^2}{2}}$ that we obtained in 4(i)
(ii) Total energy EB1
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_3}$
• Spring potential energy = 0
■ So total energy EB = $\mathbf\small{mgh_3}$
12. Now we can use the result in (8):
EA = EB1 + Energy for doing work against friction
$\mathbf\small{mgh_1+\frac{k\,x^2}{2}=mgh_3+0.392AB_1}$
$\mathbf\small{\Longrightarrow mg(h_3-h_1)=\frac{k\,x^2}{2}-0.392AB_1}$
13. From fig.6.60(b), we have: (h3-h1) = B1C1.
Also, $\mathbf\small{AB_1=\frac{B_1 C_1}{\sin 60}}$
14. So the result in (12) becomes:
$\mathbf\small{mg(B_1 C_1)=\frac{k\,x^2}{2}-0.392 \times \frac{B_1 C_1}{\sin 60}}$
• Substituting the known values, we get:
$\mathbf\small{0.2 \times 9.8(B_1 C_1)=\frac{1400\times 0.1^2}{2}-0.392 \times \frac{B_1 C_1}{\sin 60}}$
$\mathbf\small{\Longrightarrow 1.96(B_1 C_1)=7-0.392 \times \frac{B_1 C_1}{\sin 60}}$
• Thus we get B1C1 = 2.9 m
• So $\mathbf\small{AB_1=\frac{B_1 C_1}{\sin 60}=\frac{2.9}{\sin 60}=3.348\,\,\text{m}}$
Solved example 6.25
A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 Nm-1 as shown in fig. 6.61(a) below:
The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless. [g = 10 ms-2]
Solution:
1. Let 'A' be the point of release of the block
'A' is at a height of 'h1' from the datum. This is shown in fig.b
2. After stretching the spring, the block comes to rest at 'B'
'B' is at a height of 'h2' from the datum
3. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_1}$
• Spring potential energy = 0
■ So total energy EA = $\mathbf\small{mgh_1}$
(ii) Total energy EB
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_2}$
• Spring potential energy = $\mathbf\small{\frac{k\,x^2}{2}}$
■ So total energy EB = $\mathbf\small{mgh_2+\frac{k\,x^2}{2}}$
4. If there is no friction, EB must be equal to EA
• But since there is friction, EB will be less than EA.
• This is because, some energy is lost for doing work against friction
• If we add this 'lost energy' to EB, the energies will balance. That is:
EA = EB + Energy for doing work against friction
5. So our next task is to find the 'lost energy'
• For that, we have to first find the frictional force
• We have already learnt to find that in the case of inclined planes (Details here)
• So frictional force = μk × FN = (μk × mgcosθ) = (μk × 1 × 10 × cos 37) = 7.986μk N
• So energy lost = Frictional force × Distance = 7.986μk× 0.1 = 0.7986 μk N
6. Now we can use the expression in (4):
EA = EB + Energy for doing work against friction
That is: $\mathbf\small{mgh_1=mgh_2+\frac{k\,x^2}{2}+0.7986 \mu_k}$
$\mathbf\small{mg(h_1-h_2)=\frac{k\,x^2}{2}+0.7986 \mu_k}$
7. Draw a vertical through A
• Also draw a horizontal through B
• These vertical and horizontal will meet at a point. Let us call it 'C'. This is shown in fig.c
• From the right triangle ABC, we have: (h1-h2) = AC
• Also, $\mathbf\small{AC=AB\, \sin 37}$
= 0.1 × 0.60182 = 0.0602
8. So the result in (6) becomes:
$\mathbf\small{mg(0.0602)=\frac{k\,x^2}{2}+0.7986 \mu_k}$
Substituting the known values, we get:
$\mathbf\small{1 \times 10 \times(0.0602)=\frac{100\times 0.1^2}{2}+0.7986 \mu_k}$
$\mathbf\small{\Rightarrow 0.602=0.5+0.7986 \mu_k}$
$\mathbf\small{\Rightarrow \mu_k=0.1277}$
Solved example 6.26
In the fig.6.62(a) below, the two masses A and B are released from rest.
Find an expression for the velocity of the masses at the instant when the masses have traveled a distance of 'h' m. Use both force method and energy method
Solution:
• When the masses are released, B travels downwards with an acceleration
• Since the string is inextensible, A travels upwards with the same acceleration
• Because of the acceleration, the velocity goes on increasing
• We are asked to find the velocity at the instant when the masses have traveled 'h' m
Force method:
1. Fig.b shows the FBD of A
Taking upward forces as positive and downward forces as negative, we get: T-mAg = mAa
(Where 'a' is the acceleration with which the masses move)
2. Fig.c shows the FBD of B
From this we get: T-mBg = -mBa
⇒ T = mBg - mBa
Substituting this in (1), we get: mBg - mBa - mAg = mAa
⇒ mBg - mAg = mAa + mBa
⇒ (mB-mA)g = (mB+mA)a
$\mathbf\small{\Rightarrow a=\left(\frac{m_B-m_A}{m_B+m_A}\right)g}$
3. We can use the equation: $\mathbf\small{v^2-u^2=2as}$
• Substituting the known values, we get: $\mathbf\small{v^2-0^2=2ah}$
$\mathbf\small{\Rightarrow v=\sqrt{2ah}}$
Where $\mathbf\small{a=\left(\frac{m_B-m_A}{m_B+m_A}\right)g}$
Energy method:
1. The initial elevations of A and B are shown in fig.d
Let us write the energies:
(i) Total energy EA1:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{m_Agh_{A1}}$
■ So total energy EA1 = $\mathbf\small{m_Agh_{A1}}$
(ii) Total energy EB1:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{m_Bgh_{B1}}$
■ So total energy EB1 = $\mathbf\small{m_Bgh_{B1}}$
2. The final elevations (after travelling 'h' m) of A and B are shown in fig.e
Let us write the energies:
(i) Total energy EA2:
• Kinetic energy = $\mathbf\small{\frac{m_A\,v^2}{2}}$
♦ Where 'v' is the velocity with which the blocks move when they just pass the elevations shown in fig.e
• Gravitational potential energy = $\mathbf\small{m_Agh_{A2}}$
■ So total energy EA2 = $\mathbf\small{\frac{m_A\,v^2}{2}+m_Agh_{A2}}$
(ii) Total energy EB2:
• Kinetic energy = $\mathbf\small{\frac{m_B\,v^2}{2}}$
• Gravitational potential energy = $\mathbf\small{m_Bgh_{B21}}$
■ So total energy EB2 = $\mathbf\small{\frac{m_B\,v^2}{2}+m_Bgh_{B2}}$
3. Total initial energy = Ei = (EA1 + EB1) = $\mathbf\small{m_Agh_{A1}+m_Bgh_{B1}}$
• Total final energy = Ef = (EA2 + EB2) = $\mathbf\small{\frac{m_A\,v^2}{2}+m_Agh_{A2}+\frac{m_B\,v^2}{2}+m_Agh_{B2}}$
4. Applying the Law of conservation of energy, we have: Ei = Ef.
• Thus we get: $\mathbf\small{m_Agh_{A1}+m_Bgh_{B1}=\frac{m_A\,v^2}{2}+m_Agh_{A2}+\frac{m_B\,v^2}{2}+m_Bgh_{B2}}$
$\mathbf\small{\Rightarrow m_Bgh_{B1}-m_Bgh_{B2}=\frac{m_A\,v^2}{2}+m_Agh_{A2}-m_Agh_{A1}+\frac{m_B\,v^2}{2}}$
$\mathbf\small{\Rightarrow m_Bg(h_{B1}-h_{B2})=\frac{v^2}{2}(m_B+m_A)+m_Ag(h_{A2}-h_{A1})}$
• But (hB1 - hB2) = (hA2 - hA1) = h
• Thus we get: $\mathbf\small{m_Bg(h)=\frac{v^2}{2}(m_B+m_A)+m_Ag(h)}$
$\mathbf\small{(m_B-m_A)gh=\frac{v^2}{2}(m_B+m_A)}$
$\mathbf\small{\Rightarrow v^2=2 \left(\frac{m_B-m_A}{m_B+m_A}\right)gh}$
$\mathbf\small{\Rightarrow v=\sqrt{2ah}}$
• Where $\mathbf\small{a=\left(\frac{m_B-m_A}{m_B+m_A}\right)g}$
• This is the same expression that we obtained by the force method
Solved example 6.22
To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed of 18 km h-1 on a smooth road and colliding with a horizontally mounted spring of spring constant 6.25 × 103 Nm-1.
1. What is the maximum compression of the spring?
2. What will be the maximum compression if friction is also taken into account. Coefficient of kinetic friction is 0.5
Solution:
1. We know that 36 kmh-1 is 10 ms-1
$\mathbf\small{\because 36 \times \left( \frac{1000}{60 \times 60} \right)=10}$
• 18 is half of 36. So 18 kmh-1 is 5 ms-1
2. Fig.6.57 below shows the simulation:
 |
| Fig.6.57 |
• After compression of the spring, the velocity of the car becomes zero
• That means, after compression, the kinetic energy is zero
3. But the initial kinetic energy is not lost.
• Based on the law of conservation of energy, the initial kinetic energy is transformed into another form of energy: The spring potential energy
4. Now we can write the equations:
• Initial kinetic energy = $\mathbf\small{\frac{mv^2}{2}=\frac{1000 \times 5^2}{2}\,\,J}$
• Let ‘x’ be the compression of the spring
• Then potential energy acquired by the spring = $\mathbf\small{\frac{kx^2}{2}=\frac{6250 \times x^2}{2}\,\,J}$
5. Equating the two energies we get: $\mathbf\small{\frac{1000 \times 25}{2}=\frac{6250 \times x^2}{2}\,\,J}$
• Thus we get x = 2 m (This is the answer for part 1)
6. When friction is also present, we cannot just equate the two energies as in (5)
• The original energy will be the same $\mathbf\small{\frac{1000 \times 5^2}{2}\,\,J}$
• But some energy will be lost for doing work against friction
• So the car will not be able to make the same compression
7. Let x1 be the new compression
• Work done against friction during the travel through this 'x1' m
= frictional force × distance = $\mathbf\small{\mu_k \times mg \times x_1=0.5 \times 1000 \times 10 \times x_1=5000x_1 \,\, J}$
8. If we add this energy to the right side, the energies will balance. So we get:
$\mathbf\small{\frac{1000 \times 5^2}{2}=\frac{6250 \times x_1^2}{2}+5000x_1}$
$\mathbf\small{\Longrightarrow 25000=6250x_1^2+10000x_1}$
$\mathbf\small{\Longrightarrow 25=6.250x_1^2+10x_1}$
$\mathbf\small{\Longrightarrow 6.250x_1^2+10x_1-25=0}$
Solving this quadratic equation, we get:
x1= 1.35 m (This is the answer for part 2)
Another approach using force method:
Part 1:
1. The car makes initial contact with the spring with a velocity of u = 5 ms-1
• The final velocity of the car v = 0
• The acceleration 'a' experienced by the car = resistive force⁄mass
2. Resistive force is the force exerted by the spring
♦ But such a force is not uniform
♦ Because it depends on the compression 'x'
3. So we take the average force:
$\mathbf\small{\frac{0+kx}{2}=0.5kx =0.5 \times 6250 \times x=3125x \,\,\text{N} }$
• So a = $\mathbf\small{\frac{3125x}{1000}=3.125x\,\,\text{ms}^{-2}}$
• Note that, this is a negative acceleration
4. We can use the equation: $\mathbf\small{v^2-u^2=2as}$
• Substituting the known values, we get: $\mathbf\small{0^2-5^2=2 \times (-3.125x) \times x}$
$\mathbf\small{\Longrightarrow -25=-6.25x^2}$
• So x = 2.0 m
• This is the same answer that we obtained before
5. Consider the following two instants:
(i) Instant at which the car makes initial contact with the spring
(ii) Instant at which the car comes to a stop
• We want the time interval between the above two instants
6. Let us use the equation: v = u + at
• Substituting the known values, we get: 0 = 5 + (-3.125x)t
⇒ 0 = 5 - 6.25t
⇒ t = 0.8 seconds
7. While using the 'energy method', we are not able to find 'time'
• We are now able to find 'time' because we applied Newton's second law and the equations of motion
Part 2:
1. The car makes initial contact with the spring with a velocity of u = 5 ms-1
• The final velocity of the car v = 0
• The acceleration 'a' experienced by the car = resistive force⁄mass
2. Resistive force = the force exerted by the spring + frictional force
• We have already calculated the force exerted by the spring as 3125x1
• Frictional force = μkmg = 0.5×1000×10 = 5000 N
• So total resistive force = (3125x1+5000) N
• So negative acceleration = Force⁄mass = $\mathbf\small{\frac{3125x_1+5000}{1000}}$
4. We can use the equation: $\mathbf\small{v^2-u^2=2as}$
• Substituting the known values, we get: $\mathbf\small{0^2-5^2=2 \times -1 \times(\frac{3125x_1+5000}{1000}) \times x_1}$
$\mathbf\small{\Longrightarrow -25000=-6250x_1^2-10000x_1}$
$\mathbf\small{\Longrightarrow 6250x_1^2+10000x_1 -25000=0}$
$\mathbf\small{\Longrightarrow 6.25x_1^2+10x_1-25=0}$
• Solving this quadratic equation, we get:
x1= 1.35 m
• This is the same answer that we obtained before
5. Consider the following two instants:
(i) Instant at which the car makes initial contact with the spring
(ii) Instant at which the car comes to a stop
• We want the time interval between the above two instants
6. Let us use the equation: v = u + at
• Substituting the known values, we get: $\mathbf\small{0=5+-1 \times (\frac{3125x_1+5000}{1000})t_1}$
$\mathbf\small{\Longrightarrow 0=5+-1 \times (\frac{3125 \times 1.35+5000}{1000})t_1}$
$\mathbf\small{\Longrightarrow 0=5+-1 \times (\frac{4218.75+5000}{1000})t_1}$
$\mathbf\small{\Longrightarrow 5000=9218.75t_1}$
So t1 = 0.5423 seconds
• Note that, when 'resistance to motion' increases, the 'time required to bring the car to a stop' decreases
7. While using the 'energy method', we are not able to find 'time'
• In 'force method', we are able to find 'time' because we apply Newton's second law and the equations of motion
Solved example 6.23
A block of mass 8 kg is released from the top of an inclined smooth surface. See fig.6.58(a) below:
 |
| Fig.6.58 |
Solution:
1. Let 'A' be the point of release of the block
'A' is at a height of 'h1' from the datum. This is shown in fig.b
2. After compressing the spring, the block comes to rest at 'B'
'B' is at a height of 'h2' from the datum
3. By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EB
4. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_1}$
• Spring potential energy = 0
■ So total energy EA = $\mathbf\small{mgh_1}$
(ii) Total energy EB
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_2}$
• Spring potential energy = $\mathbf\small{\frac{k\,x^2}{2}}$
■ So total energy EB = $\mathbf\small{mgh_2+\frac{k\,x^2}{2}}$
5. Equating the two energies we get:
$\mathbf\small{mgh_1=mgh_2+\frac{k\,x^2}{2}}$
$\mathbf\small{\Longrightarrow mg(h_1-h_2)=\frac{k\,x^2}{2}}$
• Substituting the known values, we get: $\mathbf\small{8\times 10(h_1-h_2)=\frac{200 \times \,1^2}{2}}$
• Thus we get: (h1-h2) = 1.25 m
6. Draw a vertical through A
• Also draw a horizontal through B
• These vertical and horizontal will meet at a point. Let us call it 'C'
• A, B and C forms a right triangle as shown in fig.c
• Obviously, (h1-h2) = AC = 1.25 m
• Also, angle at B = 30o
7. In the triangle ABC, sin 30 = 1.25⁄AB
So we get: AB = 1.25⁄sin 30 = 1.25⁄0.5 = 2.5 m
Solved example 6.24
In fig.6.59(a) below, the spring constant of the spring is 1400 Nm-1. It is compressed by 0.1 m using the block and then released.
 |
| Fig.6.59 |
1. There is no friction
2. If the coefficient of kinetic friction μk = 0.4
[g = 9.8 ms-2]
Solution:
1. Let 'A' be the point of release of the block
• 'A' is at a height of 'h1' from the datum. This is shown in fig.b
2. After release, the block comes to rest at 'B'
• 'B' is at a height of 'h2' from the datum
3. By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EB
4. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_1}$
• Spring potential energy = $\mathbf\small{\frac{k\,x^2}{2}}$
■ So total energy EA = $\mathbf\small{mgh_1+\frac{k\,x^2}{2}}$
(ii) Total energy EB
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_2}$
• Spring potential energy = 0
■ So total energy EB = $\mathbf\small{mgh_2}$
5. Equating the two energies we get:
$\mathbf\small{mgh_1+\frac{k\,x^2}{2}=mgh_2}$
$\mathbf\small{\Longrightarrow mg(h_2-h_1)=\frac{k\,x^2}{2}}$
• Substituting the known values, we get: $\mathbf\small{0.2\times 9.8(h_2-h_1)=\frac{1400 \times \,0.1^2}{2}}$
• Thus we get: (h2-h1) = 3.571 m
6. Draw a vertical through A
• Also draw a horizontal through B
• These vertical and horizontal will meet at a point. Let us call it 'C'
• A, B and C forms a right triangle as shown in fig.c
• Obviously, (h2-h1) = BC = 3.571 m
• Also, angle at A = 60o
7. In the triangle ABC, sin 60 = 3.571⁄AB
So we get: AB = 3.571⁄sin 60 = 4.124 m (This is the answer for part 1)
8. If there is no friction, EB must be equal to EA
• But since there is friction, EB will be less than EA.
• This is because, some energy is lost for doing work against friction
• If we add this 'lost energy' to EB, the energies will balance. That is:
EA = EB + Energy for doing work against friction
9. So our next task is to find the 'lost energy'
• For that, we have to first find the frictional force
• We have already learnt to find that in the case of inclined planes (Details here)
• So frictional force = μk × FN = (μk × mgcosθ) = (0.4 × 0.2 × 9.8 × cos 60) = 0.392 N
10. So energy lost = Frictional force × Distance
• This time, the block will not reach up to 'B' which we saw in the fig.6.59(b) above
• Let us call the new point 'B1' This is shown in fig.6.60(a) below:
 |
| Fig.6.60 |
• Thus 'energy lost' = 0.392AB1
11. Next we write the values of EA and EB
(Note that, EB is now EB1)
(i) The position 'A' has not changed. So EA is the same $\mathbf\small{mgh_1+\frac{k\,x^2}{2}}$ that we obtained in 4(i)
(ii) Total energy EB1
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_3}$
• Spring potential energy = 0
■ So total energy EB = $\mathbf\small{mgh_3}$
12. Now we can use the result in (8):
EA = EB1 + Energy for doing work against friction
$\mathbf\small{mgh_1+\frac{k\,x^2}{2}=mgh_3+0.392AB_1}$
$\mathbf\small{\Longrightarrow mg(h_3-h_1)=\frac{k\,x^2}{2}-0.392AB_1}$
13. From fig.6.60(b), we have: (h3-h1) = B1C1.
Also, $\mathbf\small{AB_1=\frac{B_1 C_1}{\sin 60}}$
14. So the result in (12) becomes:
$\mathbf\small{mg(B_1 C_1)=\frac{k\,x^2}{2}-0.392 \times \frac{B_1 C_1}{\sin 60}}$
• Substituting the known values, we get:
$\mathbf\small{0.2 \times 9.8(B_1 C_1)=\frac{1400\times 0.1^2}{2}-0.392 \times \frac{B_1 C_1}{\sin 60}}$
$\mathbf\small{\Longrightarrow 1.96(B_1 C_1)=7-0.392 \times \frac{B_1 C_1}{\sin 60}}$
• Thus we get B1C1 = 2.9 m
• So $\mathbf\small{AB_1=\frac{B_1 C_1}{\sin 60}=\frac{2.9}{\sin 60}=3.348\,\,\text{m}}$
Solved example 6.25
A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 Nm-1 as shown in fig. 6.61(a) below:
 |
| Fig.6.61 |
Solution:
1. Let 'A' be the point of release of the block
'A' is at a height of 'h1' from the datum. This is shown in fig.b
2. After stretching the spring, the block comes to rest at 'B'
'B' is at a height of 'h2' from the datum
3. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_1}$
• Spring potential energy = 0
■ So total energy EA = $\mathbf\small{mgh_1}$
(ii) Total energy EB
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh_2}$
• Spring potential energy = $\mathbf\small{\frac{k\,x^2}{2}}$
■ So total energy EB = $\mathbf\small{mgh_2+\frac{k\,x^2}{2}}$
4. If there is no friction, EB must be equal to EA
• But since there is friction, EB will be less than EA.
• This is because, some energy is lost for doing work against friction
• If we add this 'lost energy' to EB, the energies will balance. That is:
EA = EB + Energy for doing work against friction
5. So our next task is to find the 'lost energy'
• For that, we have to first find the frictional force
• We have already learnt to find that in the case of inclined planes (Details here)
• So frictional force = μk × FN = (μk × mgcosθ) = (μk × 1 × 10 × cos 37) = 7.986μk N
• So energy lost = Frictional force × Distance = 7.986μk× 0.1 = 0.7986 μk N
6. Now we can use the expression in (4):
EA = EB + Energy for doing work against friction
That is: $\mathbf\small{mgh_1=mgh_2+\frac{k\,x^2}{2}+0.7986 \mu_k}$
$\mathbf\small{mg(h_1-h_2)=\frac{k\,x^2}{2}+0.7986 \mu_k}$
7. Draw a vertical through A
• Also draw a horizontal through B
• These vertical and horizontal will meet at a point. Let us call it 'C'. This is shown in fig.c
• From the right triangle ABC, we have: (h1-h2) = AC
• Also, $\mathbf\small{AC=AB\, \sin 37}$
= 0.1 × 0.60182 = 0.0602
8. So the result in (6) becomes:
$\mathbf\small{mg(0.0602)=\frac{k\,x^2}{2}+0.7986 \mu_k}$
Substituting the known values, we get:
$\mathbf\small{1 \times 10 \times(0.0602)=\frac{100\times 0.1^2}{2}+0.7986 \mu_k}$
$\mathbf\small{\Rightarrow 0.602=0.5+0.7986 \mu_k}$
$\mathbf\small{\Rightarrow \mu_k=0.1277}$
Solved example 6.26
In the fig.6.62(a) below, the two masses A and B are released from rest.
 |
| Fig.6.62 |
Solution:
• When the masses are released, B travels downwards with an acceleration
• Since the string is inextensible, A travels upwards with the same acceleration
• Because of the acceleration, the velocity goes on increasing
• We are asked to find the velocity at the instant when the masses have traveled 'h' m
Force method:
1. Fig.b shows the FBD of A
Taking upward forces as positive and downward forces as negative, we get: T-mAg = mAa
(Where 'a' is the acceleration with which the masses move)
2. Fig.c shows the FBD of B
From this we get: T-mBg = -mBa
⇒ T = mBg - mBa
Substituting this in (1), we get: mBg - mBa - mAg = mAa
⇒ mBg - mAg = mAa + mBa
⇒ (mB-mA)g = (mB+mA)a
$\mathbf\small{\Rightarrow a=\left(\frac{m_B-m_A}{m_B+m_A}\right)g}$
3. We can use the equation: $\mathbf\small{v^2-u^2=2as}$
• Substituting the known values, we get: $\mathbf\small{v^2-0^2=2ah}$
$\mathbf\small{\Rightarrow v=\sqrt{2ah}}$
Where $\mathbf\small{a=\left(\frac{m_B-m_A}{m_B+m_A}\right)g}$
Energy method:
1. The initial elevations of A and B are shown in fig.d
Let us write the energies:
(i) Total energy EA1:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{m_Agh_{A1}}$
■ So total energy EA1 = $\mathbf\small{m_Agh_{A1}}$
(ii) Total energy EB1:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{m_Bgh_{B1}}$
■ So total energy EB1 = $\mathbf\small{m_Bgh_{B1}}$
2. The final elevations (after travelling 'h' m) of A and B are shown in fig.e
Let us write the energies:
(i) Total energy EA2:
• Kinetic energy = $\mathbf\small{\frac{m_A\,v^2}{2}}$
♦ Where 'v' is the velocity with which the blocks move when they just pass the elevations shown in fig.e
• Gravitational potential energy = $\mathbf\small{m_Agh_{A2}}$
■ So total energy EA2 = $\mathbf\small{\frac{m_A\,v^2}{2}+m_Agh_{A2}}$
(ii) Total energy EB2:
• Kinetic energy = $\mathbf\small{\frac{m_B\,v^2}{2}}$
• Gravitational potential energy = $\mathbf\small{m_Bgh_{B21}}$
■ So total energy EB2 = $\mathbf\small{\frac{m_B\,v^2}{2}+m_Bgh_{B2}}$
3. Total initial energy = Ei = (EA1 + EB1) = $\mathbf\small{m_Agh_{A1}+m_Bgh_{B1}}$
• Total final energy = Ef = (EA2 + EB2) = $\mathbf\small{\frac{m_A\,v^2}{2}+m_Agh_{A2}+\frac{m_B\,v^2}{2}+m_Agh_{B2}}$
4. Applying the Law of conservation of energy, we have: Ei = Ef.
• Thus we get: $\mathbf\small{m_Agh_{A1}+m_Bgh_{B1}=\frac{m_A\,v^2}{2}+m_Agh_{A2}+\frac{m_B\,v^2}{2}+m_Bgh_{B2}}$
$\mathbf\small{\Rightarrow m_Bgh_{B1}-m_Bgh_{B2}=\frac{m_A\,v^2}{2}+m_Agh_{A2}-m_Agh_{A1}+\frac{m_B\,v^2}{2}}$
$\mathbf\small{\Rightarrow m_Bg(h_{B1}-h_{B2})=\frac{v^2}{2}(m_B+m_A)+m_Ag(h_{A2}-h_{A1})}$
• But (hB1 - hB2) = (hA2 - hA1) = h
• Thus we get: $\mathbf\small{m_Bg(h)=\frac{v^2}{2}(m_B+m_A)+m_Ag(h)}$
$\mathbf\small{(m_B-m_A)gh=\frac{v^2}{2}(m_B+m_A)}$
$\mathbf\small{\Rightarrow v^2=2 \left(\frac{m_B-m_A}{m_B+m_A}\right)gh}$
$\mathbf\small{\Rightarrow v=\sqrt{2ah}}$
• Where $\mathbf\small{a=\left(\frac{m_B-m_A}{m_B+m_A}\right)g}$
• This is the same expression that we obtained by the force method
So we have completed a discussion on conservation of mechanical energy. Next we will see the law of conservation in other forms of energy such as heat energy, sound energy, nuclear energy etc.,
