Part 21 (1/2)

CHAPTER VI

WORK AND ENERGY

_104. Work._--”Whenever a force moves a body upon which it acts, it is said to do work upon that body.” For example, if a man pushes a wheelbarrow along a path, he is doing work on it as long as the wheelbarrow moves, but if the wheelbarrow strikes a stone and the man continues to push and no motion results, from a scientific point of view he is then doing no work on it.

”Work signifies the overcoming of resistance,” and unless the resistance is overcome no work is done. Lifting a weight is doing work on it, supporting a weight is not, although the latter may be nearly as tiresome as the former. Work as used in science is a technical term. Do not attach to it meanings which it has in every-day speech.

=105. Measurement of Work.=--Work is measured by the product of the force by the displacement caused in the direction of the force, that is _W_ = _fs_. Therefore if a unit of force acts through a unit of s.p.a.ce, a unit of work will be done. There are naturally several units of work depending upon the units of force and s.p.a.ce employed.

_English Work Unit._--If the force of one _pound_ acts through the distance of one _foot_, a _foot-pound_ of work is done. A foot-pound is defined as the work done when 1 lb. is lifted 1 ft. against the force of gravity.

_Metric Work Unit._--If the force is one _kilogram_ and the distance one _meter_, _one kilogram-meter_ of work is done.

_Absolute Work Unit._--If the force of one _dyne_ acts through the distance of one _centimeter_ a _dyne-centimeter_ of work is done. This usually is called an i. Other work units are sometimes used depending upon the force and distance units employed. One, the i, is equal to 10,000,000 ergs or 107 ergs.

=Problem.=--If a load is drawn 2 miles by a team exerting 500 lbs.

force, how much work is done?

=Solution.=--Since the force employed is 500 lbs., and the distance is 2 5280 ft., the work done is 500 2 5280 or 5,280,000 ft.-lbs.

=106. Energy.=--In the various cases suggested in the paragraphs upon work, an agent, a man, an animal or a machine, was mentioned as putting forth an effort in order to do the work. It is also true that in order to perform work an agent must employ _energy, or the energy of a body is its capacity for doing work_. Where an agent does work upon a body, as in winding up a spring or in lifting a weight, the body upon which the work has been done may acquire energy by having work done upon it. That is, it may become able to do work itself upon some other body. For instance, a lifted weight in falling back to its first position may turn wheels, or drive a post into the ground against resistance; a coiled spring may run clock work, strike a blow, or close a door. Hence the energy, or the capacity for doing work, is often acquired by a body because work has first been done upon that body.

=107. Potential Energy.=--The wound up spring may do work because work has first been done upon it. The lifted weight may also do work because work has first been done in raising it to its elevated position since in falling it may grind an object to powder, lift another weight or do some other kind of work. _The energy that a body possesses on account of its position or shape and a stress to which it is subjected is called potential energy._ The potential energy of a body is measured by the work done in lifting it, changing its shape, or by bringing about the conditions by which it can do work. Thus if a block of iron weighing 2000 lbs. is lifted 20 ft., it possesses 40,000 ft.-lbs. of potential energy. It is therefore able to do 40,000 ft.-lbs. of work in falling back to its first position. If the block just mentioned should fall from its elevated position upon a post, it could drive the post into the ground because its motion at the instant of striking enables it to do work. To compute potential energy you compute the work done upon the body. That is, _P.E._ = _w_ _h_ or _f_ _s_.

=108. Kinetic Energy.=--_The energy due to the motion of a body is called kinetic energy_. The amount of kinetic energy in a body may be measured by the amount of work done to put it in motion. It is usually computed, however, by using its ma.s.s and velocity on striking. To ill.u.s.trate, a 100-lb. ball is lifted 16 ft. The work done upon it, and hence its potential energy, is 1600 ft.-lbs. On falling to the ground again, this will be changed into kinetic energy, or there will be 1600 ft.-lbs. of kinetic energy on striking. It will be noted that since energy is measured by the work it can do, work units are always used in measuring energy. To compute the kinetic energy of a falling body by simply using its ma.s.s and velocity one proceeds as follows, in solving the above problem:

First, find the velocity of the falling body which has fallen 16 ft. A body falls 16 ft. in _one_ second. In this time it gains a velocity of 32 ft. per second. Now using the formula for kinetic energy _K.E._ = _wv_/(2_g_), we have _K.E._ = 100 32 32/(2 32) = 1600 ft.-lbs. as before. The formula, _K.E._ = _wv_/(2_g_), may be derived in the following manner:

The kinetic energy of a falling body equals the work done in giving it its motion, that is, _K.E._ = _w_ _S_, in which, _w_ = the weight of the body and _S_ = the distance the body must fall freely in order to acquire its velocity. The distance fallen by a freely falling body, _S_, = 1/2_gt_ = _g__t_/(2_g_) (Art. 98, p. 111).

Now, _v_ = _gt_ and _v_ = _g__t_.

Subst.i.tuting for _g__t_, its equal _v_, we have _S_ = _v_/(2_g_). Subst.i.tuting this value of S in the equation _K.E._ = _w_ _S_, we have _K.E._ = _wv_/(2_g_).

Since the kinetic energy of a moving body depends upon its ma.s.s and velocity and not upon the _direction_ of motion, this formula may be used to find the kinetic energy of any moving body. Ma.s.s and weight in such problems may be considered numerically equal.

=Important Topics=

1. Work defined.

2. Work units, foot-pound, kilogram-meter, erg.

3. Energy defined.

4. Kinds of energy, potential and kinetic.

=Problems=

1. How much work will a 120-lb. boy do climbing a mountain 3000 ft.

high? Should the vertical or slant height be used? Why?