我算到一半就做不出來了,麻煩大大們教一下!!!!!!lol....
A car having a mass of 1000 kilograms starts from rest from the top of a hill 100 meters high. At the bottom of the hill the car is moving 30 m/s. The energies involved in this situation are kinetic , potential (mgh), and friction (label it FR).
a) Write an equation involving the energies involved.
b) Determine the energy due to friction.
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a)
TOTAL ENERGY = sum of all energies involved
Kinetic: + ½ mv^2=450000
Potential: + mgh hint here: 0
Friction: - FR <= 從這開始我就卡住了!!!!
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b)
m = 1000 kg
h = 100 m
Initial speed = 0 m/s
Final speed = 30 m/s....然後我不知從何下手*0*
(( 急!!!)) potential, friction and kinetic energy!!!!
2008-10-23 11:35 am
回答 (2)
2008-10-24 5:27 am
✔ 最佳答案
(a) The equation is:loss of potential energy at top of hill = gain in kinetic energy at bottom of hill + work done against friction
i.e. mgh = (1/2).m.v^2 + W
where W is the work done against friction
m is the mass of the car
h is the height of the hill, and
v is the speed of the car at bottom of the hill
g is the acceleration due to gravity
(b) Substituting the numerical values into the above equation,
1000g x 100 = (1/2) x 1000 x 30^2 + W
i.e. W = [1000g x 100 - (1/2) x 1000 x 30^2] J
2008-10-23 6:03 pm
Energy may not conservate here.
Take g=10m/s^2, on top of hill
E=PE+KE=PE=1000000J (no v)
At the bottom
E=PE+KE=KE=1/2 mv^2=450000J
Different = 1000000-450000=550000J
This is the lost due to friction.
The energy due to friction: =f . s (friction dot displacement), that is the same as the work done by a force. But in here, no information of that was provide.
Take g=10m/s^2, on top of hill
E=PE+KE=PE=1000000J (no v)
At the bottom
E=PE+KE=KE=1/2 mv^2=450000J
Different = 1000000-450000=550000J
This is the lost due to friction.
The energy due to friction: =f . s (friction dot displacement), that is the same as the work done by a force. But in here, no information of that was provide.
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081023000051KK00196