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Course by: Sunil Kumar Singh. E-mail the author

# Conservation of energy in mechanical process

Module by: Sunil Kumar Singh. E-mail the author

Summary: Conservation of energy in mechanical process is a subset of the law of conservation of energy.

We have discussed formulation of conservation of energy in general. In this module, we shall discuss conservation law as applied to mechanical process. Typically, a mechanical process involves three different kinds of energy - kinetic, potential and heat. We shall, however, restrict treatment of heat as an equivalent to work by friction.

On the other hand, the process generally involves gravitational, elastic and friction forces. These forces transfer energy involving different energy forms by doing work on the objects of the system.

Depending on the context of situation, we use either closed or isolated system. Further, mechanical process is generally limited to exchange of energy within or outside the system via “work”. In short, we can describe a mechanical process in terms of following elements :

• System : closed or isolated
• Energy : kinetic, potential and heat (work by friction)
• Forces : gravitational, elastic and friction
• Transfer of energy : work by forces across and within the system

In order to facilitate systematic study of mechanical process, we shall study three different situations in progressive simplicity of the process :

1. A mechanical process with external force on the system.
2. A mechanical process without external force on the system.
3. A mechanical process without external force on the system and without non-conservative internal force (absence of friction).

In this module, we shall discuss the first two processes. Third process represents the ideal mechanical process. The corresponding conservation law is known as “conservation of mechanical energy”. This law is being dealt in a separate module.

## Mechanical process with external force on the system

One implication of external force on the system is that we are dealing with special “closed” system, which permits exchange of energy with surrounding via “work by external force” only. The form of conservation law for the mechanical process is :

W E = Δ E mech + Δ E thermal W E = Δ E mech + Δ E thermal

However, we pointed out that we shall limit change in thermal energy to “heat energy”, which is equal to the negative of work by friction only. Hence, we can rewrite the equation as :

W E = Δ E mech W F W E = Δ E mech W F

W E + W F = Δ E mech W E + W F = Δ E mech

W E + W F = Δ K + Δ U W E + W F = Δ K + Δ U

In words, we can put the conservation of energy for mechanical process as :

“Work by external force on the system and work by friction within the system is equal to the change in potential and kinetic energy of the system.”

### Example

Problem 1: Two blocks of 1 kg and 2 kg are connected by a spring as shown in the figure. If spring constant is 500 N/m and coefficient of static and kinetic friction between surfaces are each 0.5, then what minimum constant horizontal force, F, (in Newton) is required to just initiate block on the left ? (Consider g = 10 m / s 2 m / s 2 ).

Solution : The system consists of two blocks and one spring. The system is subjected to an external force. Thus, system is “closed” system, which allows energy exchange with system via work by external force. As friction is also involved, the corresponding energy statement is :

W E + W F = Δ K + Δ U W E + W F = Δ K + Δ U

We are required to find minimum force. We need to understand the implication of this phrase. It is clear that we can apply external force in such a manner that block of “2 kg” acquires kinetic energy by the time block of “1 kg” is initiated in motion. Alternatively as a base case, we can apply external force gradually and slowly in increasing magnitude till the block of “1 kg” is initiated in motion. In this case, the block of “2 kg” does not acquire kinetic energy. This mode of application of external force represents the situation when minimum external force will be required to initiate block of “1 kg”. Hence,

Δ K = 0 Δ K = 0

Motion of block is constrained in horizontal direction. There is no change of vertical elevation. Hence, there is no change in gravitational potential energy. However, spring is stretched from its neutral state. As a consequence, there is change in elastic potential energy of the spring. Let “x” be the extension in the spring by the time block of “1 kg” is initiated in motion. Then, the total change in potential energy is :

Δ U = 1 2 k x 2 Δ U = 1 2 k x 2

The work by friction is done only on the right block of “2 kg” :

W F = - μ m 2 g x W F = - μ m 2 g x

On the other hand, the work by external force is :

W E = F x W E = F x

Putting all these values in the equation of conservation of energy :

F x μ m 2 g x = 1 2 k x 2 F x μ m 2 g x = 1 2 k x 2

F μ m 2 g = 1 2 k x F μ m 2 g = 1 2 k x

Clearly, we can not solve this equation as there are two unknowns, “F” and “x”. We, therefore, make use of the fact that spring force on the block of mass “1 kg” is equal to maximum static friction,

k x = μ m 1 g k x = μ m 1 g

Combining two equations, we have :

F μ m 2 g = 1 2 μ m 1 g F μ m 2 g = 1 2 μ m 1 g

F = μ g m 1 + m 2 2 = 0.5 X 10 1 + 2 2 X 10 = 10 N F = μ g m 1 + m 2 2 = 0.5 X 10 1 + 2 2 X 10 = 10 N

#### Note:

It is interesting to note that force required to initiate left block is independent of spring constant.

## Mechanical process without external force on the system

Since no external force operates on the system, there is no work by external force. The system under consideration is, therefore, an isolated system. The form of conservation law for general mechanical process is further reduced as :

W F = Δ K + Δ U W F = Δ K + Δ U

In words, we can put the conservation of energy for mechanical process under given condition as :

“Work by friction within an isolated system is equal to the change in potential and kinetic energy of the system.”

### Example

Problem 2: In the arrangement shown, the block of mass 10 kg descends through a height of 1 m after being released. The coefficient of friction between block and the horizontal table is 0.3, whereas pulley is friction-less. Considering string and pulley to be “mass-less”, find the speed of the blocks.

Solution : Here, friction is present as internal force to the system. Hence, we use the form of conservation law as :

W F = Δ K + Δ U W F = Δ K + Δ U

Let us denote 6 kg and 10 kg blocks with subscript “1” and “2”. There is no change of height for the block on the table. Only change in gravitational potential energy is due to change in the height of the block hanging on the other end of the string. Thus, change in potential energy is :

Δ U = m 1 g h + m 2 g h = 0 10 x 10 x 1 = - 100 J Δ U = m 1 g h + m 2 g h = 0 10 x 10 x 1 = - 100 J

Two blocks are constrained by a taut string. It means that both blocks move with same speed. Let the speed of the blocks after traveling “1 m” be "v". Now, initial kinetic energy of the system is zero. Therefore, the change in kinetic energy is given by :

Δ K = 1 2 m 1 v 2 + 1 2 m 2 v 2 Δ K = 1 2 m 1 v 2 + 1 2 m 2 v 2

Δ K = 1 2 X 6 X v 2 + 1 2 X 10 X v 2 Δ K = 1 2 X 6 X v 2 + 1 2 X 10 X v 2

Δ K = 8 v 2 Δ K = 8 v 2

Friction works only on the block lying on the table. Here, work by friction is given as :

W F = - μ N x = - μ m 1 g x = - 0.3 X 6 X 10 X 1 = - 18 J W F = - μ N x = - μ m 1 g x = - 0.3 X 6 X 10 X 1 = - 18 J

Putting in the equation of conservation of energy,

- 18 = 8 v 2 100 - 18 = 8 v 2 100

v 2 = 82 8 v 2 = 82 8

v = 3.2 m / s v = 3.2 m / s

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