# How to find the energy of an object that was at rest?

#### haruspex

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the change in the velocity of the second rocket, which was $1.01v_i$?
That is not the change in velocity.
why did you said that it is rather inaccurate?
What is √1.04?

#### haruspex

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It is the new velocity.
Right, so what is the change in velocity?
It is 1.0198...
So what would be the right way to express that to two decimal places?

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#### Davidllerenav

Right, so what is the change in velocity?
it would be $\Delta v= v_f-v_0$ so the change would be $\Delta v= 1.012v_0-v_0=0.012v_0$, right?

Right, so what is the change in velocity?

So what would be the right way to express that to two decimal places?
1.012?

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#### haruspex

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it would be $\Delta v= v_f-v_0$ so the change would be $\Delta v= 1.012v_0-v_0=0.12v_0$, right?

1.012?
No, 1.0198 would not be abbreviated to 1.012; and even if it were, subtracting 1 does not yield 0.12.

#### Davidllerenav

No, 1.0198 would not be abbreviated to 1.012.
Then what would it be?
and even if it were, subtracting 1 does not yield 0.12.
Yes sorry, I meant 0.012.

#### haruspex

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Then what would it be?
You will kick yourself when you realise the error you are making.
Repeatedly increment the final nonzero digit of 1.0198 until there are only two decimal places.

#### Davidllerenav

Repeatedly increment the final nonzero digit of 1.0198 until there are only two decimal places.
Oh, I think I see it know, woudl it be 1.02?

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#### Davidllerenav

So $\Delta v=0.02$ and the kinetic energy of the second rocket, the one that was at rest is $E_k=m\Delta v^2= m(0.02)^2v_0^2$? Is t right?

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#### Davidllerenav

Not quite... you forgot a factor.
Oh, yes the $\frac{1}{2}$, so it is $E_k=\frac{1}{2}m\Delta v$.

#### haruspex

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Oh, yes the $\frac{1}{2}$, so it is $E_k=\frac{1}{2}m\Delta v$.
Now you forgot the square.

#### Davidllerenav

Now you forgot the square.
Sorry... $E_k=\frac{1}{2}m(\Delta v)^2= \frac{1}{2}m(0.02)^2v_0^2=\frac{1}{2}m(0.004)v_0^2$. It think it is correct know.

#### haruspex

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Sorry... $E_k=\frac{1}{2}m(\Delta v)^2= \frac{1}{2}m(0.02)^2v_0^2=\frac{1}{2}m(0.004)v_0^2$. It think it is correct know.
I'm sorry to say that's another arithmetic error.

#### Davidllerenav

I'm sorry to say that's another arithmetic error.
I see, I made a mistake when I squared 0.02, it is 0.0004, not 0.004. so $E_k=\frac{1}{2}m(0.0004)v_0^2. #### haruspex Science Advisor Homework Helper Gold Member 2018 Award I see, I made a mistake when I squared 0.02, it is 0.0004, not 0.004. so$E_k=\frac{1}{2}m(0.0004)v_0^2$. Right. And you can simplify slightly combining the 1/2 with the 4. #### Davidllerenav Right. And you can simplify slightly combining the 1/2 with the 4. Yes, so simplifying I end up with$E_k=m(0.0002)v_0^2$. And that's all, right? #### haruspex Science Advisor Homework Helper Gold Member 2018 Award Yes, so simplifying I end up with$E_k=m(0.0002)v_0^2$. And that's all, right? Perfect. #### Davidllerenav Ok, thanks. I don't need to write the equations of the momentum of both rocket, do I? #### haruspex Science Advisor Homework Helper Gold Member 2018 Award Ok, thanks. I don't need to write the equations of the momentum of both rocket, do I? No, you already used that. #### Davidllerenav #### jbriggs444 Science Advisor Homework Helper Yes, so simplifying I end up with$E_k=m(0.0002)v_0^2$. And that's all, right? There is a problem with this result. Neither$m$nor$v_0$are given in the problem description. Instead, what is given is$E_0$, the initial energy of the moving rocket. Can you express the result you have in terms of$E_0$rather than in terms of$m$and$v_0$? #### Davidllerenav There is a problem with this result. Neither$m$nor$v_0$are given in the problem description. Instead, what is given is$E_0$, the initial energy of the moving rocket. Can you express the result you have in terms of$E_0$rather than in terms of$m$and$v_0$? Well, since$E_0=\frac{1}{2}m v_0^2$, I think I can express the result as$E_k=(0.0004)E_0$right? #### jbriggs444 Science Advisor Homework Helper Well, since$E_0=\frac{1}{2}m v_0^2$, I think I can express the result as$E_k=(0.0004)E_0## right?
Yes, that is what I get as well.

"How to find the energy of an object that was at rest?"

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