CHAPTER XV. ENERGY. 170. DEF. Energy is capacity for doing work. There are many different forms of energy, of which two chiefly concern us in Dynamics; viz. Potential Energy and Kinetic Energy. 171. DEF. Potential energy is capacity for doing work which a mass has by virtue of its position. A spring when bent, has energy by reason of its position; for it can do work when changing that position. Air when compressed can do work on being allowed to expand. A mass at the top of a tower can do work in its descent to a lower level. Example. Find the difference between the Potential Energy of 1 cwt. at the top of a tower 100 ft. high, and that of 1 cwt. at the foot of the tower. In descending from the top of the tower to the bottom, the weight of the I cwt. does 100 × 112 foot-pounds; Hence, I cwt. at the top of a tower 100 ft. high, has 100 × 112 footpounds more Potential Energy than I cwt. at the foot of the tower. 172. DEF. Kinetic energy is the capacity for doing work which a mass has by virtue of its mass-velocity. A bullet when in motion can do work in giving up its velocity. Water when in motion can do work in turning a water wheel. The mass of a hammer can do work by virtue of its velocity. The kinetic energy of a particle having a given velocity is measured by the number of foot-poundals it does in giving up that velocity. 173. PROP. Prove that the difference between the energy of m lbs. having v velos, and that of m lbs. having u velos, is (1⁄2mv2 – 1⁄2mu3) foot-poundals. Let a force of p poundals be applied to the m lbs. so that in seconds its velocity is reduced from v velos to u velos; also suppose that in that interval the mass has a retardation a and passes over s feet in the direction opposite to the force; then, by Art. 26. iii, as = {v2 - u2, or, mas = {m22 – } mu2; but ma is the number of poundals required to produce in m lbs. acceleration a; so that ma=p. But ps is the number of foot-poundals done by poundals which acts on a mass while it passes over s feet in its own direction. That is to having velocity v can do (m2 while its velocity is reduced from v to u. 174. say, the mass m lbs. mu2) foot-poundals Q.E.D. Hence The kinetic energy of a particle of m lbs. which has v velos is mv2 foot-poundals. For the mass can do mv2 foot-poundals against a force before it is reduced to rest. Also, if m2 foot-poundals of work are done on a particle of mass m lbs. at rest, it will have v velos. Example. Find the work done by the engine of a train of 120 tons, moving the train a mile from rest and at the same time giving it a speed of 30 miles per hour; the motion being on a horizontal line and the resistance due to friction of the weight of the train. The work done against friction is (2 × 2240×120) × (1760 × 3) foot-pounds, =1120 × 1760 × 3 foot-pounds. is The work done on the inertia, in producing kinetic energy, × 120 × 2240 × (44)2 foot-poundals Hence, the whole work done by the engine is (1120 × 30 × 176 + 1120 × 30 × 2 × 121) foot-pounds EXAMPLES. XXXIX. Find the kinetic energy of each of the six following; 1. A stone of 4 lbs. which has 200 velos. 2. A cannon shot of 8 cwt. which has 2000 velos. 3. A cannon shot of 12 cwt. which has 1600 velos. [Art. 173.] 4. A train of 400 tons going at the rate of 60 miles an hour. 5. A man of 12 stone running at the rate of 12 miles an hour. 6. A man of 12 stone after falling a distance of 10 feet. 7. Find the work done by the engine of a train of 360 tons, in going 2 miles from rest on horizontal rails, the friction being of the weight, and the speed attained 45 miles per hour. 8. Find the work necessary to raise a train of 100 tons to the top of a mountain pass 500 feet high and at the same time to give it a velocity of 60 miles an hour (neglecting friction etc.). 9. A man has to raise I cwt. of bricks 8 feet; he throws them up so that they arrive at a point 8 ft. high with velos; compare the superfluous work done with the necessary work. 10. A tram car of 1 ton is stopped by a brake 10 times in going a mile; the brake stops the car in II yards; after each stoppage the car attains a velocity of 7 miles an hour. Supposing the friction of the rails to be a uniform force of 28 lbs. weight, compare the work done in this journey with the work done in going a mile with uniform velocity. 11. The expense of moving a train is proportional to the work done. Compare the cost of getting the speed of a train up from rest to a velocity of 45 miles an hour and at the same time going I mile, with the cost of moving it a mile with uniform velocity; supposing that the resistance caused by friction etc. to be of the weight of the train." 12. Suppose the tram-car in Question 9 had no brake, and so had to be stopped by the horses; find how much more work the horses would have to do in consequence in a journey which includes 10 stoppages. 175. When a particle slides along a smooth plane the pressure of the plane, whatever it is, is always perpendicular to the plane, and therefore perpendicular to the path of the particle; the distance through which the pressure works is therefore zero; hence 176. The pressure of a smooth plane on a particle moving along its surface does no work on the particle. Example. When a particle slides down a smooth inclined plane under the action of gravity only, its total energy is unchanged. By Example i. p. 91, a particle of m lbs. sliding down an inclined plane of angle a has g sin a celos. Let h feet be the vertical distance passed over by the particle in any interval seconds during which the velocity of the particle is changed from u velos to v velos. Then the particle has lost mgh foot-poundals of potential energy and gained (mv2 - mu2) foot-poundals of kinetic energy; we have to prove that these amounts of work are equal. then but Let s feet be the distance passed over on the plane, Therefore 177. The sum of the potential energy and of the kinetic energy of a particle is unaltered by any motion under the action of gravity only. Example i. Let m lbs. fall from rest h feet vertically. Then the potential energy is diminished by mh foot-pounds. It receives v velos, where mgh=m2; so that it has received kinetic energy capable of doing mgh foot-poundals, that is, mh foot-pounds; in other words, the potential energy lost is equal to the kinetic energy gained. ་ Example ii. Since the velocity of a projectile is equal to that due to a fall from the directrix, it follows that (the kinetic energy+the potential energy) of a projectile is always equal to the potential energy of an equal particle at rest at some point on the directrix. 178. When a particle slides along a curved surface which is smooth, it is understood that the pressure of the surface on the particle is always perpendicular to the surface, and therefore this pressure is always perpendicular to the path of the particle. The distance through which the pressure works is therefore always zero; so that the work done by the pressure is zero; hence, 179. When a particle is sliding on a smooth surface under the action of no forces its kinetic energy is constant; that is, its speed is constant. *180. We may also prove this important proposition as follows: [See also Ex. 21, p. 144.] PROP. An inelastic particle slides along the smooth sides of a regular polygon; prove that, in the limit when the polygon becomes a circle, the impacts at the angular points have no effect on its speed. Let AB, BC be two of the sides of the regular polygon. Produce AB to K. Let the velocity along AB be v velos. Resolve this velocity along and perpendicular to BC; the resolved parts are v cos a velos and v sin a velos. The resolved part v sin a velos is destroyed by the impact. Hence after impact the particle will have v cos a velos along BC. Similarly after the next impact it will have (v cos a) cos a velos along the next side and so on. Hence, when the velocity of the particle is turned through a finite angle A, where na = A, it will have v cos"a velos, that is n ข (cos 4) velos. |