A pendulum bob of mass 0.01kg is pulled to one side and raised through a vertical height 0.1m.
C)If the bob is given a push so that it has an initial speed of 1ms^-1,what is its height above its lowest point at the other end?
I think the solution is KE+PE=KE 2+PE 2,but it has 2 unknown.AM i think wrongly??
0.02h+0.005=0.005v,Dont just give me the answer,i want the reason,
Phy Question*Mechanics
2008-01-13 9:02 pm
回答 (2)
2008-01-13 9:18 pm
✔ 最佳答案
KE1 + PE1 = PE2 (All KE and PE are converted to PE when the bob attains its maximum height)1/2mu^2 + mgh1 = mgh2
1/2 (1)^2 + 10(0.1) = 10h2
h2 = 0.15m
2008-01-13 13:22:54 補充：
Or you may sayWhen the bob attains the maximum height, the bob is momentarily at rest.i.e. v = 0 == KE = 0
2008-01-13 9:30 pm
the solution is KE+PE=KE 2+PE 2 ....
but this situation 1 and 2 should be the initial one and the final one ...
I guess your situation 2 is when the bob is at the lowest point ....
but here you do not have the v ... so you should not use this
Instead, you should use the point where the bob is pushed
Thus,
mgh = mg(0.1) + m(1)^2 / 2
gh = 0.1 g + 0.5
h = 1.5 m
but this situation 1 and 2 should be the initial one and the final one ...
I guess your situation 2 is when the bob is at the lowest point ....
but here you do not have the v ... so you should not use this
Instead, you should use the point where the bob is pushed
Thus,
mgh = mg(0.1) + m(1)^2 / 2
gh = 0.1 g + 0.5
h = 1.5 m
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