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PROBLEM 1.13\n\nWeight refers to the force of gravity; F_grav = mg\nThus, when her mass is 120 lb and weight is 119.16 lb, we have\n\ng = F_grav / m = (119.16 lb) / (120 lb) = 0.992 ft/s² = 31.9606 ft/s²\n\nWhen her mass is 120 lb and g = 32.05 ft/s², we have\n\nF_grav = m*g = (120 lb)(32.05 ft/s²)\n= 1 lb\n= 11.954 lb\n\nCOMMENT: Her mass remains constant, but weight depends on the local acceleration of gravity.\n\nPROBLEM 1.14\n\nThe actual forces developed when birds and aircraft collide are difficult to determine precisely, but estimates can be calculated using average values of acceleration and force magnitudes, as follows:\n\nThe gull is accelerated from a very low velocity, to 150 miles/h in 10^-5 s. Thus the average acceleration magnitude is\n\n|a| = (150 miles/h - 0) / (2600 ft / 1 mile) * (1 ft / 10^-5 s)\n= 2.2 x 10^5 ft/s²\n\nThe magnitude of the average force applied is\n\n|F| = m|a| = (12.1 lb)(2.2 x 10^5 ft/s²) / 32.2 lb / ft/s² = 82,000 lb (rounded)\n\nPROBLEM 1.15\n\nm = 4.5 lb\na = 3g where g = 32.2 ft/s²\n∑ F_z = ma\nNeglecting air resistance,\nF - F_grav = ma\n=> F = ma + F_grav = m(a + g)\n= m(4g)\n= (4.5 lb)(4 x 32.2 ft/s²) / 32.2 lb /ft/s² = 181 lb (rounded)