I ton = 2240 lbs., and 5 yards = 15 feet. Expressed in foot-pounds, the work done is 2240 x 15 3360. Expressed in foot-poundals, the work is 2240 x 15 x 32 = 107520 (taking g = 32). 75. The weight of a tram-car is 8 tons, and the resistance due to friction encountered in moving it is equal to one-sixteenth of the weight of the car: how much work is done in a run of 4 miles? The resistance to motion weight of ton, = = weight of 1120 pounds. The distance through which this force is overcome is 4 miles = 4 × 1760 × 3 feet, =21120 feet. The work done (expressed in gravitation-units) is 1120 × 21120 foot-pounds = 23,654,400 foot-pounds. 76. Assuming that a person walking on level ground does work equivalent to the raising of his own weight vertically upwards through one-twentieth of the distance walked, find (in foot-tons) the average daily work done by Weston in his walk of 5000 miles in 100 days, his weight being 9 stone 2 lbs. The average daily walk was 50 miles, and the average daily work was equivalent to the raising of his own weight through Hence the of The weight raised was 9 stone 2 lbs. = 128 lbs. 50 x 88 × 3 × 128 - 5280754 II2 X 20 7 77. A mass of 12 kilogrammes is raised through a vertical height of 8 metres: express the work done in muammo contimetres and convert this into eras 78. A man can pump 30 gallons of water per m to a height of 16 feet: how many foot-pounds of does he do in an hour? 79. An agent A exerts a force equal to a weig 50 lbs. through a distance of 120 feet, and another B exerts a force equal to a weight of 180 lbs. th 90 feet. What is the ratio between A's work and 80. A ladder 20 feet long rests against a vertica and is inclined at an angle of 30° to it: how much is done by a man weighing 13 stone in ascending i 81. A body of mass 3 lbs. is projected vert upwards with a velocity of 640 feet per second: much work has been done against gravity when i ascended to half its maximum height? 82. How much work would be done in lifting 8 grammes to a height of 12 metres above the surfa the moon, where g is 150? 83. Two masses M and M1 are acted upon b same force for the same time: find the relation bet (1) the amounts of momentum generated in the ma (2) the amounts of work done upon them. 84. A body of mass 12 lbs. rests upon a horiz plane, the coefficient of friction between it and the I being 0.14: find the work done in moving the through a distance of 4 yards along the plane. 85. If the plane in the preceding question inclined at an angle of 30° to the horizontal, how I work would have to be done in order to move the 3 yards along it? 86. Calculate, in foot-tons, the work done in mc a railway train weighing 120 tons through a distan 2 miles along a level line, assuming that the resista amount to 12 lbs. for every ton in motion. 87. If we change from a foot-pound-second yard-pound-minute system, in what ratio must we the unit of work? 88. Find the work done in drawing a carriage c tons up an incline one mile in length and rising 1 in 120, the coefficient of friction being 110. 89. The cylinder of a steam-engine has a diameter of 6 inches, and the piston moves through a distance of 10 inches find the work done per stroke, assuming the pressure of the steam in the cylinder to be constant and equal to 30 lbs. per square inch. 90. Two bodies of 80 lbs. and 60 lbs. are raised through heights of 100 feet and 50 feet respectively. Calculate the total amount of work done, and show that it is equal to the work which would be done in raising the sum of the weights through a vertical distance equal to that through which their centre of gravity is raised. 91. Assuming the result indicated in the preceding question, and taking the weight of one cubic foot of water as 62.5 lbs.; find how much work must be done in order to empty a well 10 feet in diameter and 200 feet deep, filled to the brim with water. Energy. The energy of a body is the power which it possesses of doing work, When it possesses this power in virtue of its position (as when it is raised above the level of the ground) the energy is called statical or potential energy. When it possesses the power of doing work in virtue of its motion, the energy is called kinetic energy (K.E.) The weight of a body of mass m grammes is mg dynes; if the body be raised to a height of h centimetres above the level of the ground, the work which it can do in falling is mgh ergs, or The potential energy of a body of mass m, raised to a height h, is migh. If m is expressed in pounds and h in feet, the product mgh will be the measure of the energy in foot-poundals; but expressed in foot-pounds the measure of the energy will be mh simply. the unit of kinetic energy is that possessed by two of mass moving with unit velocity (not that posse by unit mass moving with unit velocity.)] 92. A reservoir contains water at a height of 200 above the ground: what is the potential energy of water in foot-pounds per gallon? The potential energy of each pound of water in the rese is 200 foot-pounds, and one gallon of water = 10 lbs. H the potential energy per gallon is 10 x 200=2000 foot-po 93. What is the potential energy of a mass of kilogrammes raised to a height of 40 metres above ground? Its energy = 25 x 40 kilogramme-metres, = 1000 x 105 gramme-centimetres, 94. A stone of mass 6 kilogrammes falls from res a place where g=980: what will be its kinetic ene at the end of 5 seconds? ds with a have have to h; and e of the Ossessed elocity v > units sessed o feet f the ervoir Ience unds. f 25 : the : at roy The velocity acquired in 5 seconds will be v=gx5=980 x 5 = 4900, and since 6 kilos. = 6000 gm., = K.E. x 6000 × (4900)2=7.203 × 1010 ergs. 95. What is the K.E. of a body of mass 16 lbs. moving with a velocity of 50 feet per second? Expressed in foot-poundals the K.E. of the body is ×16× (50)2 = 20,000. Since one foot-pound = 32 foot-poundals, the K. E. in footpounds is [We might have commenced by defining kinetic energy as being the value of the product mv2; then proceeding, as follows, to show that this quantity is a measure of the work which a body can do in virtue of its motion : Suppose a body of mass m moving with a velocity u to be acted upon by a force F in the direction of its motion, and let the acceleration produced by this force After the body has moved through a space s will be given by the equation F m be a= But since F=ma, mas = Fs, and Fs is the work done by the force F acting through the space s. Thus the work done by the force is measured by the increase of kinetic energy which it produces; or, [K. E. at any time] = [initial K.E.] + [work done by acting force]. If the force acts upon the body in a direction opposite |