Work and Power

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 Work:W = F x d = Fd cosq 1.How much work does Fred do exerting 45 N to lift a box 3.2 m? W = F x d W = (45)(3.2) W = 144 J 2.What distance will 90 J of energy slide a box against 12 N of frictional force? W = F x d 90 = (12)d 90s = 12 d = 90/12 d = 7.5 m 3.Katherine moves a box 7.2 m doing 9 J of work.What is the fricitonal force? W = F x d 9 = F(7.2) F = 9/7.2 F = 1.25 N Work and weight:DEp = mgDh 4.How much work is needed to lift a 45 Kg box 9.2 m? DEp = mgDh DEp = (45)(9.8)(9.2) DEp = 4057.2 DEp = 4060 J 5.A pump does 4500 J of work in a minute.What mass of water can it pump to a height of 4.5 m in one minute? DEp = mgDh 4500 = m(9.8)(4.5) m = 4500/(9.8)(4.5) m = 102 Kg 6.A rifle cartridge contains 250 J of energy.How far can it launch its 60 gram slug into the air if it is shot straight up? DEp = mgDh 250 = (60 g)(9.8)h 250 = (.06 Kg)(9.8)h 250 = (.588)h h = 250/.588 h = 425 m Power:P = DW/Dt 7.A motor can do 540 J of work in 9.0 seconds.What's its power output? P = DW/Dt P = 540/9 P = 60 W 8.In what time can a 250 W motor do 4500 J of work? P = DW/Dt 250 = 4500/Dt Dt = 4500/250 Dt = 18 s 9.How much work does a 25 W motor do in 5 minutes? P = DW/Dt 25 = DW/5 min 25 = DW/(5 min)(60 s/1 min) 25 = DW/300 DW = (25)(300) DW = 7500 J Jambalaya:W = F x s,DEp = mgDh,P = DW/Dt 10.What must be the power rating of a motor that can lift a 600 Kg load 30 meters in 5 seconds? DEp = mgDh DEp = (600)(9.8)(30) DEp = 176400 DEp = DW P = DW/Dt P = 176400/5 P = 35280 W 11.A sled dog drags a 240 Kg sled 50 m in 35 seconds when the coefficient of friction between the snow and the runners is .056.What is the power output of the dog? W = F x d,P = DW/Dt,Ffr£mkFN,FN = mg Ffr£mkFNFN = mg Ffr£ (.056)(240)(9.8) Ffr£ 131.712 W = F x d W = (131.712)(50) W = 6565.6 P = DW/Dt P = 6585.6/35 P = 188 W 12.Greg LeMond can put out ¾ of a horsepower.(1 HP = 745.7 W).In what time can he climb a 2000 m mountain if he and his bike have a mass of 82 Kg? (Ignore Friction) DEp = mgDh DEp = (82)(9.8)(2000) DEp = 1607200 J DEp = DW (3/4)(745.7) = 559.275 W P = DW/Dt 560 = 1607200/Dt Dt = 1607200/560 Dt = 2870 s 13.My van can only go about 25 m in one second on a level road at full power.(30 HP,1 HP = 745.7 W)What must be the force of friction opposing my van?(most of this is air friciton and drag) W = F x s,P = DW/Dt P = (30)(745.7) = 22371 W P = DW/Dt 22371 = DW/1 W = F x d 22371 = F(25) F = 900 N 14.A steam engine must drag logs across 45 m of level rock.if the coefficient of friction between the logs and the ground is .78, and the engine can put out 5 Horse power (1 HP = 745.7 W), what is the maximum mass of logs it can drag in 15 seconds? W = F x d,P = DW/Dt,Ffr£mkFN,FN = mg P = (5)(745.7) = 3728.5 3728.5 = DW/15 DW = (3728.5)(15) DW = 55927.5 W = F x d 55927.5 = F(45) F = 55927.5/45 F = 1241.8 Ffr£mkFN,FN = mg 1241.8 £ (.78)(9.8)(m) m = 163 kg 15.A hiker can put out 250 W of power.If they have a mass of 72 Kg and are carrying a 43 Kg pack, what elevation gain can they expect to achieve in 6 hours of hiking? P = DW/Dt DEp = mgDh DEp = DW P = mgDh/Dt 250 = (72 + 43)(9.8)h/6 hr 250 = (115)(9.8)h/(6 hr)(60 min/1 hr)(60 s/1 min) 250 = (1127)h/21600 s 5400000 = 1127h h = 5.4 x 106/1127 h = 4791.48 h = 4800 m