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 mass and velocity on striking. To illustrate, 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 mass 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. = wv2/(2g), we have K.E. = 100 × 32 × 32/(2 × 32) = 1600 ft.-lbs. as before. The formula, K.E. = wv2/(2g), 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, = 1/2gt2 = g2t2/(2g) (Art. 98, p. 111). Now, v = gt and v2 = g2t2.
Substituting for g2t2, its equal v2, we have S = v2/(2g). Substituting this value of S in the equation K.E. = w × S, we have K.E. = wv2/(2g).
Since the kinetic energy of a moving body depends upon its mass and velocity and not upon the direction of motion, this formula may be used to find the kinetic energy of any moving body. Mass and weight in such problems may be considered numerically equal.