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Can you figure this out!? |
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| Jan5-05, 08:39 AM | #1 |
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Can you figure this out!?
Suppose that a cart is being moved by a certain net force. If the net force is doubled, by how much does the acceleration change? Why does this happen? For what reason?
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| Jan5-05, 08:52 AM | #2 |
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[tex] F=ma [/tex] and taking into account that mass wouldn't change,what do you think it will happen to the acceleration? Daniel. |
| Jan5-05, 08:58 AM | #3 |
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The "rigorous" proof should follow from an analysis of the equations [itex]F_1=ma_1[/itex] and [itex]F_2=ma_2[/itex]. What is [itex]a_2[/itex] in terms of [itex]a_1[/itex] if you set [itex]F_2=2F_1[/itex]?
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| Jan5-05, 09:00 AM | #4 |
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Can you figure this out!?
The acceleration slows, right? Why does this happen?
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| Jan5-05, 09:02 AM | #5 |
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Wouldn't u agree?? Daniel. |
| Jan5-05, 09:04 AM | #6 |
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I have no idea what you are talking about quasar! Could you explain?
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| Jan5-05, 09:05 AM | #7 |
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I do agree thanks! Any you explain quasar987 reply?
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| Jan5-05, 09:11 AM | #8 |
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Let's say u have a body of mass "m".U apply a force on it.Call it "F".The second law of dynamics says that the acceleration imprimed by this force (call it "a") is nothing but
[tex] a=\frac{F}{m} [/tex] Now apply the force doubled.Which means the force 2F.Call the new acceleration "a'" ("a" prime)?Again,the second law says that the acceleration is the ratio between force and mass [tex] a'=\frac{2F}{m}=2\frac{F}{m}=2a [/tex] ,where u made use of the first formula to express the new acceleration in terms of the old one. Therefore,the acceleration doubles. Daniel. |
| Jan5-05, 09:15 AM | #9 |
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I get it now! thank you for your help!
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