206_PartUniversity Physics Solution

206_PartUniversity - Work and Kinetic Energy 6.86 Pav = F!vav Use F = ma to calculate the force IDENTIFY SET UP 6-23 vav = 0 6.00 m/s = 3.00 m/s 2

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Work and Kinetic Energy 6-23 6.86. IDENTIFY: av av PF v = ! . Use Fm a = to calculate the force. SET UP: av 06 . 0 0 m / s 3.00 m/s 2 v + == EXECUTE: Your friend&s average acceleration is 2 0 6.00 m/s 2.00 m/s 3.00 s vv a t = . Since there are no other horizontal forces acting, the force you exert on her is given by 2 net (65.0 kg)(2.00 m/s ) 130 N a = . av (130 N)(3.00 m/s) 390 W P . EVALUATE: We could also use the work-energy theorem: 2 1 21 2 (65.0 kg)(6.00 m/s) 1170 J WK K =−= = . av 1170 J 390 W 3.00 s W P t = , the same as obtained by our other approach. 6.87. IDENTIFY: To lift a mass m a height h requires work Wm g h = . To accelerate mass m from rest to speed v requires 2 1 2 WK K m v . av W P t Δ = Δ . SET UP: 60 s t = EXECUTE: (a) 25 (800 kg)(9.80 m/s )(14.0 m) 1.10 10 J (b) (1/2)(800 kg)(18.0 m/s ) 1.30 10 J. (c) 55 1.10 10 J 1.30 10 J 3.99 kW. 60 s ×+ × = EVALUATE: Approximately the same amount of work is required to lift the water against gravity as to accelerate it to its final speed. 6.88. IDENTIFY: v = ! and a = ! . SET UP: From Problem 6.71, 2 23 vtt αβ =+ and 2 6 at α β = + . EXECUTE: 222 2 3 (2 6 )(2 3 ) (4 18 18 ) vm a β tt β tmt β t β t αα = + + = + + ! . 22 33 (0.96 N/s) (0.43 N/s ) (0.043 N/s ) Pt t t + . At 4.00 s, t = the power output is 13.5 W. EVALUATE: P increases in time because v increase and because a increases. 6.89. IDENTIFY and SET UP: Energy is av . The total energy expended in one day is the sum of the energy expended in each type of activity. EXECUTE: 4 1 day 8.64 10 s Let walk t be the time she spends walking and other t be the time she spends in other activities; 4 other walk . The energy expended in each activity is the power output times the time, so 7 walk other (280 W) (100 W) 1.1 10 J EP t t t + 47 walk walk (280 W) (100 W)(8.64 10 s ) 1.1 10 J = × 6 walk (180 W) 2.36 10 J t 4 walk 1.31 10 s 218 min 3.6 h. t = = EVALUATE: Her average power for one day is 7 (1.1 10 J)/([24][3600 s]) 127 W. ×= This is much closer to her 100 W rate than to her 280 W rate, so most of her day is spent at the 100 W rate. 6.90. IDENTIFY and SET UP: WP t = EXECUTE: (a) The hummingbird produces energy at a rate of 0.7 J/s to 1.75 J/s. At 10 beats/s, the bird must expend between 0.07 J/beat and 0.175 J/beat. (b) The steady output of the athlete is (500 W)/(70 kg) 7 W/kg, = which is below the 10 W/kg necessary to stay aloft. Though the athlete can expend 1400 W/70 kg 20 W/kg = for short periods of time, no human-powered aircraft could stay aloft for very long. EVALUATE: Movies of early attempts at human-powered flight bear out our results. 6.91. IDENTIFY and SET UP: Use Eq.(6.15). The work done on the water by gravity is mgh , where 170 m. h = Solve for the mass m of water for 1.00 s and then calculate the volume of water that has this mass.
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6-24 Chapter 6 EXECUTE: The power output is 9 av 2000 MW 2.00 10 W. P == × av W P t Δ = Δ and 92% of the work done on the water by gravity is converted to electrical power output, so in 1.00 s the amount of work done on the water by gravity is 9 9 av (2.00 10 W)(1.00 s) 2.174 10 J 0.92 0.92 Pt W Δ× , Wm g h = so the mass of water flowing over the dam in 1.00 s must be 9 6 2 1.30 10 kg (9.80 m/s )(170 m) W m gh × = × density m V = so 6 33 1.30 10 m .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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206_PartUniversity - Work and Kinetic Energy 6.86 Pav = F!vav Use F = ma to calculate the force IDENTIFY SET UP 6-23 vav = 0 6.00 m/s = 3.00 m/s 2

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