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Homework_9_solution

Homework_9_solution - Homework 9 Slides 5 EK...

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Homework 9 Slides: 5) E K = ½(1000kg)(10m/s) 2 = 50,000J , E f K - E i K = ½(1000kg)(1600m 2 /s 2 ) - ½(1000kg)(900m 2 /s 2 ) = (800,000 – 450,000)J = 350,000J , With traffic more less energy on 95 because 95 would have less traffic and less stop and go. Without traffic US1 would consume less energy because the speed would be less. Acceleration 1 = (10m/s – 0)/(10s) = 1m/s 2 ; Force 1 = (1000kg)(1m/s 2 ) = 1000N; Distance 1 = 𝜐 o 𝑡 + ½ 𝘢 t 2 = ½ (1m/s 2 )(100 s 2 ) = 50m; W = (1000N)(50m) = 50,000J , P = E/ 𝑡 = (50,000J)/10s = 5,000W Acceleration 2 = (40m/s – 30m/s)/(10s) = 1m/s 2 ; Force 2 = (1000kg)(1m/s 2 ) = 1000N; Distance 2 = 𝜐 o 𝑡 + ½ 𝘢 t 2 = (30m/s)(10s) + ½(1m/s 2 )(100s 2 ) = 350m; W = (1000N)(350m) = 350,000J , P = E/ 𝑡 = (350,000J)/10s = 35,000W 6) Velocity 1 = ( 𝜐 y 2 + 𝜐 x 2 ); 𝜐 y1 = 𝜐 yo + 𝘨 t, t = d/ 𝜐 x = (10m)/(15m/s) = (2/3)s; 𝜐 y1 = (20m/s) + (- 10m/s 2 )(2/3)s = (60/3)m/s – (20/3)m/s = (40/3)m/s; 𝜐 = [(40/3) 2 + (45/3 ) 2 ]m/s = ( 5/3 145 )m/s ; E P = 𝑚 g(h+ 𝛥 h); 𝛥 h = 𝜐 o 𝑡 + ½ 𝘢 t 2 = (20m/s)(6/9)s + ½(-10m/s 2 )(4/9)s 2 = (100/9)m; E P = (5kg)(10m/s 2 )[(100+225)/9]m = 1805.5J , E K = ½ 𝑚𝜐 2 = ½(5kg)(3625/9)m 2 /s 2 = 1007.0J , W 𝘨 = 𝛥 h( 𝑚𝘨 ) = (100/9)(5kg)(10m/s 2 ) = ( 5000/9)J Velocity 2 = 𝜐 x = 15m/s ; E P = mg(h+ 𝛥 h); 𝛥 h = ( 𝜐 f 2 + 𝜐 o 2 )/2 𝘨 = [(0m/s) 2 + (20m/s) 2 ]/2(10m/s 2 ) = 20m; h = 25m + 𝛥 h = 45m; E P = (5kg)(10m/s 2 )(45m) = 2250J , E K = ½ 𝑚𝜐 2 = ½(5kg)(15m/s) 2 = 562.5J , W 𝘨 = 𝛥 h( 𝑚𝘨 ) = (20m)(5kg)(10m/s 2 ) = 2250J Velocity 3 = ( 𝜐 y 2 + 𝜐 x 2 ); 𝜐 y1 = 𝜐 yo + 𝘨 t, t = d/ 𝜐 x – t 2 = (60m)/(15m/s) - (30m)/(15m/s)= 2s; 𝜐 y1 = (0m/s) + (-10m/s 2 )(2s) = -20m/s ; 𝜐 = [ (-20m/s) 2 + (15m/s) 2 ] = [ 625 ]m/s = 25m/s; E P = mg(h max + 𝛥 h); 𝛥 h = 𝜐 o 𝑡 + ½ 𝘢 t 2 = (0m/s)(2s) + ½(-10m/s 2 )(4s 2 ) = -20m; E P = (5kg)(10m/s 2 )(45-20)m = 1250J , E K = ½ 𝑚𝜐 2 = ½(5kg)(25m/s) 2 = 1562.5J , W 𝘨 = [(h max + 𝛥 h)- h 0 ]( 𝑚𝘨 ) = [25-(45-20)]m(5kg)(10m/s 2 ) = 0J Velocity 4 = ( 𝜐 y 2 + 𝜐 x 2 ); 𝜐 y1 = 𝜐 yo + 𝘨 t, t = d/ 𝜐 x – t 2 = (70m)/(15m/s) - (30m)/(15m/s)= (8/3)s; 𝜐 y1 = (0m/s) + (-10m/s 2 )(8/3)s = (-80/3)m/s; 𝜐 = [ ( (-80/3) 2 + (45/3 ) 2 ]m/s = (8425/9)m/s ; E P = mg(h max + 𝛥 h); 𝛥 h = 𝜐 o 𝑡 + ½ 𝘢 t 2 = (0m/s)(8/3s) + ½(-10m/s 2 )(64/9)s 2 = (-320/9)m; E P = (5kg)(10m/s 2 )(45- 320/9)m = 472.2J , E K = ½ 𝑚𝜐 2 = ½(5kg)(8425/9)m 2 /s 2 = 2340.3J , W 𝘨 = [(h max + 𝛥 h)- h 0 ]( 𝑚𝘨 ) = [(225/9)- ((405/9)+(-320/9))]m(5kg)(10m/s 2 ) = (140/9)m(5kg)(10m/s 2 ) = 777.8J

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Velocity 5 = ( 𝜐 y 2 + 𝜐 x 2 ) = 𝜐 y1 = 𝜐 yo + 𝘨
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Homework_9_solution - Homework 9 Slides 5 EK...

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