Gravitation

CONSERVATION OF MECHANICAL ENERGY

Consider a particle that is subject only to conservative forces. We may combine the work-energy theorem, W _NET = ∆ K, and the definition of potential energy, W _c = −∆ U. Since W _NET = W _c, we have, ∆ K = −∆ U or

∆ K + ∆ U = 0 (5.1 a)

Since ∆K = K _f − K _i and ∆U = U _f − U _i, this may be written as

K _f + U _f = K _i + U _i (5.1 b)

Equation 5.1 b says that although the kinetic energy and the potential energy change individually, their sum has the same value at any position. The mechanical energy E is defined as

E = K + U (5.2)

In terms of this function, Eqs. 5.1 a and 5.1 b take the forms

Conservation of mechanical energy

∆ E =0 E f = E i (5.3)

Equations 5.1 and 5.3 express the principle of the conservation of mechanical energy.

In order for us to apply the conservation of mechanical energy there should be no net work done by any external force or any internal nonconservatice force. Each potential energy function, such as U_g or U _sp, takes into account the work done by the (internal) conservative force. A second condition is that the energies must be measured in the same inertial frame. This is necessary because velocity, and hence the kinetic energy, depend on the frame of reference.

FIGURE 5.1

Let us see how the motion of an object in free-fall may be described in terms of the conservation of mechanical energy. Since U_g = mgy, the mechanical energy is

E = 1/2 mυ² + mgy

and the conservation law takes the form

+ mgy_f = + mgy_i (5.4)

Suppose that the particle starts from rest at a height H above the ground, as in Fig. 5.1 a. At this point it has no kinetic energy but has potential energy U = mgH. As it falls, its height decreases and its speed increases. That is, it loses potential energy and gains kinetic energy, but the sum, E = К + U, remains constant. Just before it lands, the object has its maximum kinetic energy, К = 1/2mυ²_MAX, and its potential energy is zero. The variation of kinetic energy and potential energy with у is shown in Fig. 5.1 b. In summary,

E = mυ² + mgy = mυ²_max = mgH (5.5)

When the object lands it experiences a force due to the ground, and so its mechan­ical energy is no longer conserved.

Дата добавления: 2014-01-05