The law of conservation of mechanical energy

January 1st, 2012

   The energy is one of the concepts in physics, for which there exists no any definition, ie, undefined concept. We can feel the energy released  in the form of heat when through a conductor pass a current , we can estimate the reserve of potential energy of the ball above the Earth’s surface according to the magnitude of its deformation after impact on the surface, etc.

   Yet energy concept is an intuitive concept.

   American physicist Richard Feynman said that if there was a man who would have given the definition of energy, humankind would have put a monument to him. Despite the fact that the energy is the undefined concept it is a physical quantity that has the remarkable property that the change in energy can be measured in different  processes (note: not the energy, but only a change!).

   Due to this property in physics emit different types of energy: the internal energy of the body, the energy of the electric field, the magnetic field energy, the mechanical energy.

   In this section, we consider the mechanical energy, its types: potential and kinetic energy, as well as one of the fundamental laws of nature –  the law of conservation of mechanical energy.

  The work of a force

   For the mathematical formulation of the law let’s consider such a thing as  the work of the force . Let the material point (abbreviation: MP) performs the movement of the point with coordinate  x_1 in a position x_2 under the action of   \vec F, which makes an angle $latex \alpha$ with a displacement vector (fig.1).

Fig.1. The calculation of the work of a force.

     Let’s perform the expansion force \vec F into two components: the normal component and the tangential component. The normal component \vec F_n is perpendicular to the x-axis and does not perform work on the displacement of the MP which is moving along the x-axis. The work on the displacement of the MP which is moving along the x-axis carries out the tangential component. \vec F_x. This work is defined as the product of the magnitude of the tangential component  and the magnitude of the displacement \Delta x=x_2-x_1:

A=F_x\Delta x=F \Delta xcos\alpha        (1)

   If the force \vec F is not constant and depends on the displacement (eg, elastic force), it is necessary to divide the movement into elementary sections  dx within which the tangential component of the \vec F_x can be considered as constant. Then the elementary work dA can be represented as:

dA=F_xdx                     (2)

Integrating the last expression we obtain  an integral representation of the work of the force:

A=\int_{x_1}^{x_2}F_xdx                                            (3)

   This expression for the work of the force will allow us to obtain an expression for the potential energy and kinetic energy at the same time as the two types of the same energy – mechanical energy.

The potential energy of a body in the gravity  field

    Let’s represent the integral (3) as the difference of two integrals:

A=\int_{0}^{x_2}F_xdx-\int_{0}^{x_1}F_xdx (4)

  Let’s introduce the notation: U = A=\int_{0}^{x}F_xdx and let’s call a scalar magnitude U as the potential energy of the body.

   Let’s use the obtained relations for the analysis of a body motion in the gravity field (Fig. 2).

Fig.2. The movement of a body under the influence of gravity .

      Let the body falls under the influence of gravity from the point with coordinate  x_1 to the point with coordinate x_2. Then, according to (4), the work that performs the force of gravity on the body movement is equal to:

A=\int_{0}^{x_2} (-mg)dx-\int_{0}^{x_1}(-mg)dx=mgx_1-mgx_2 (5)

       Or:

A=-(U2-U1)  (6)

     It can be shown that equation (6) holds not only for a rectilinear trajectory but also for the trajectory of the arbitrary shape.Let’s consider an arbitrary curved trajectory (Fig. 3).

Fig.3. Calculation of the work of the gravity force.

     Let’s partition the path on a set of mutually perpendicular segments. When a sufficiently large number n segments – so large that n tends to infinity – the trajectory that constructed by a broken line is a good approximation to a curved path. Then the work of the gravity force on the displacement of the body moving along a curved path can be a good approximation to replace the work of the same strength along the broken line.

     The work of the gravity force along the broken line is the sum of the work of the gravity force   along the straight line segments parallel to the force of gravity, and of the work of the gravity force on straight stretches perpendicular to the gravity force. The work of the gravity force on straight stretches perpendicular to the force of gravity is equal to zero in accordance with formula (1) the article “The work of a force”. Consequently, the work of the force of gravity along the broken line is the sum of the work of the gravity force along the straight line segments parallel to the force of gravity. It is clear that this sum is equal to the work of the gravity force performing the   displacement of the body moving along a straight path between the points  x_1 and x_2.

     Therefore, the work of the gravity force  does not depend on the shape of the  body trajectory. The forces that have this remarkable property, called conservative forces.

Thus, for the conservative forces have a place the equality (6).

If the work of the force depends on the shape of the trajectory, the force called the dissipative force . An example of dissipative forces can be the friction force.

     From the definition of work it consequent that the work of the force is a positive value, therefore in the formula (6), U2<U1. Consequently, the potential energy of a body falling under gravity decreases. Where does disappear this energy? The answer to this question is given by the following section:

The kinetic energy of the moving body

Let’s consider the expression dA=F_xdx and perform a series of simple transformations

dA=F_xdx=F_xvdt=\frac{d(mv)}{dt}vdt=d(mv)v=mvdv (7)

The expression (7) makes it possible to present the work in a different form:

A=\int_{x_1}^{x_2} F_xdx=\int_{v_1}^{v_2}mvdv=\frac{mv_2^2}{2}-\frac{mv_1^2}{2} (8)

The scalar value \frac{mv^2}{2}  is called  the kinetic energy . We see  from the expression (8) that  the work of the force is equal to the increase in the kinetic energy . It should be noted here that this result was obtained without specifying the type of force.

The law of a conservation of the mechanical energy

      Comparing the right and left sides of the equations (5) and (8) we see that the work of the gravity force leads to the loss of potential energy (5), at the same time the work of the force (in this case, the force of gravity) contributes to the increase of the kinetic energy:

mgx_1- mgx_2=\frac{mv_2^2}{2}-\frac{mv_1^2}{2} (9)

Let us transform this equation to the form:

mgx_1+\frac{mv_1^2}{2}= mgx_2+\frac{mv_2^2}{2} (10)

     From equation (10) it follows that at each point of the trajectory of the sum of kinetic and potential energy of the system of bodies (the body in the gravitational field and the Earth, creating the field) is constant! The body and Earth form a system of bodies, that is:

1. closed system because the bodies interact with each other only;
2. conservative system because conservative forces act each other.
These definitions allow us to formulate the law of conservation of mechanical energy in more general form:

the sum of potential and kinetic energy of a closed conservative system is constant.

TASKS

Task 1. Uniform cylinder of radius 4 cm is rolling on a horizontal surface at a speed of 3 \frac{m}{s} and reaches foundations inclined plane with an angle of 30^0 to the horizontal. Determine which the height rises the cylinder on an inclined plane on.

Task 2. The hammer weighing 400kg  falls on fixed pile weighing 100 kg at a speed of 5 \frac{m}{s}. Determine the efficiency of the hammer blow.

Task 3.  To stretch the spring by 5 cm, it is necessary to perform the work A_1 . To further stretch the spring by \Delta x cm, you need to perform additional work A_2 . Find \Delta x, if \frac{A_2}{A_1}=3.

(The information has been taken from e-book «Fundamentals of classical physics” (author V.D. Shvets), which was published in 2007 by “1C-Multimedia Ukraine” publishing company. Copyright is reserved by certificate Number 55955 from 06.08.2014.More information can be found at https://ipood-kiev.academia.edu/ValentynaShvets/Books)

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