average force
2017/07/04 09:20
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標題:
average force
發問:
Mr.Hu of mass 58 kg steps off a 10 m high platform and dives into water . Hecomes to rest when he is 4 m below the water surface . a)Estimate the speed of Mr.Hu just before he reaches the water surface . b)What is the magnitude of the average force exerted on him by the water . 更新: 請問有冇一個情況之下要加埋自身重量(即mg)? 更新 2: 我唔明 (F-mg).s = (1/2)mv^2 點解-mg b)v^2-u^2=2(a+g)s 點解加g
最佳解答:
(a) Use conservation of energy (1/2)mv^2 = mgh where m is tha mass of Hu (=58 kg) v is the speed just before entering into water g is the acceleration due to gravity h is the height of fall(=10m) (1/2)(58)v^2 = 58x10x10 v^2 = 200 (m/s)^2, thus v = 14.14 m/s (b) Use (work done) = change of kinetic energy i.e. (net force)x distance = change of kinetic energy (F-mg).s = (1/2)mv^2 where s is the distance travelled into water F is the force exerted by water (F - 58x10)x4 = (1/2)(58)(14.14^2) solve for F gives F = 2030 N
a)v^2-u^2=2gs,v=14.14ms-1 b)v^2-u^2=2(a+g)s 0-14.14^2=2(a+10)*4 a=-35ms-2 F=ma F=58*-35 F=-2030N 所以the magnitude of the average force exerted on him by the water=2030N
average force
發問:
Mr.Hu of mass 58 kg steps off a 10 m high platform and dives into water . Hecomes to rest when he is 4 m below the water surface . a)Estimate the speed of Mr.Hu just before he reaches the water surface . b)What is the magnitude of the average force exerted on him by the water . 更新: 請問有冇一個情況之下要加埋自身重量(即mg)? 更新 2: 我唔明 (F-mg).s = (1/2)mv^2 點解-mg b)v^2-u^2=2(a+g)s 點解加g
最佳解答:
(a) Use conservation of energy (1/2)mv^2 = mgh where m is tha mass of Hu (=58 kg) v is the speed just before entering into water g is the acceleration due to gravity h is the height of fall(=10m) (1/2)(58)v^2 = 58x10x10 v^2 = 200 (m/s)^2, thus v = 14.14 m/s (b) Use (work done) = change of kinetic energy i.e. (net force)x distance = change of kinetic energy (F-mg).s = (1/2)mv^2 where s is the distance travelled into water F is the force exerted by water (F - 58x10)x4 = (1/2)(58)(14.14^2) solve for F gives F = 2030 N
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其他解答:a)v^2-u^2=2gs,v=14.14ms-1 b)v^2-u^2=2(a+g)s 0-14.14^2=2(a+10)*4 a=-35ms-2 F=ma F=58*-35 F=-2030N 所以the magnitude of the average force exerted on him by the water=2030N
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