- #1

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**use the principles of conservation of energy**.

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- Thread starter dralion87
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- #1

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- #2

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Start with the definitions. The mechanical energy lost is [tex]f \Delta s=\Delta E_{therm}[/tex], where f is the magnitude of the friction force(what is the definition of kinetic friction?), and [tex]\Delta s[/tex] is the distance that you are trying to find.

By the work-energy theorem, [tex]W_{ext}=\Delta E_{mech}+\Delta E_{therm}[/tex]. Now, solve for [tex]\Delta s[/tex]

- #3

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- #4

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- #5

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im lost really i just dont get it..would you kbe kind enough and plug in nummbers

- #6

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- #7

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kinetic energy is (1/2) M V^2 and F= friction

- #8

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No, I meant force of friction. What is the definition of force of friction? After you've done that, then you can plug it in the equation that I told you:

[tex]W_{ext}=\Delta E_{mech}+\Delta E_{therm}[/tex]

You know,

(1) [tex] f \Delta s=\Delta E_{therm} [/tex]

(2) [tex]W_{ext}=\Delta U + \Delta K + f\Delta s[/tex]

(2) As you've just said, [tex] \Delta K =\frac{1}{2}mv^2[/tex]

Now, ask yourself, what must [tex]W_{ext}[/tex] be? Then write the definition of f. Then find s.

[tex]W_{ext}=\Delta E_{mech}+\Delta E_{therm}[/tex]

You know,

(1) [tex] f \Delta s=\Delta E_{therm} [/tex]

(2) [tex]W_{ext}=\Delta U + \Delta K + f\Delta s[/tex]

(2) As you've just said, [tex] \Delta K =\frac{1}{2}mv^2[/tex]

Now, ask yourself, what must [tex]W_{ext}[/tex] be? Then write the definition of f. Then find s.

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- #9

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this is wat i have

(1/2)(1500)v^2+.05 delta s =.05 delta s

(1/2)(1500)v^2+.05 delta s =.05 delta s

- #10

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Where did you get the terms on the right-hand side? What is [tex]W_{ext}[/tex]? Hint: There are no conservative forces and no external forces acting on the system, then W=?

- #11

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- #12

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[tex]W=\frac{1}{2}mv^2+\mu mg \Delta s[/tex]

You're almost there. What must W be? Then you can solve for [tex]\Delta s[/tex]

- #13

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w must be the same as (1/2)mv^2 + "mu"mg

- #14

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- #15

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it should equal 0

- #16

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it should equal 0

- #17

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Yes! Now, solve for the distance, [tex] \Delta s [/tex]

- #18

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i got 1.02m i dont think it my answer makes sense

- #19

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What do you mean it doesn't make sense? Show your work.

- #20

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alright this is wat i put in.... (1/2)(1500)(cos180)^2 divided by (.05)(1500)(9.8)

- #21

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That makes no sense whatsover, plug W=0 back into the equation.

[tex]W=0=\frac{1}{2}mv^2+\mu mg \Delta s[/tex]

Solve for [tex] \Delta s[/tex] I fail to see the issue here.

[tex]W=0=\frac{1}{2}mv^2+\mu mg \Delta s[/tex]

Solve for [tex] \Delta s[/tex] I fail to see the issue here.

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- #22

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is v^2 equal to cos 180.... or what do i pug in for velocity since its not given

- #23

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Obviously, you won't get a numerical value and [tex]v^2 \neq cos(180)[/tex](where are you getting this from? this is nonsense). You are not given the numerical value of the initial velocity, so of course, you won't get a numerical value for the distance.

Look, use your ALGBEBRA skills to solve for [tex] \Delta s [/tex]. Anything else, you're wasting your time.

You are close, you have the [tex]\Delta x[/tex] as some constant * [tex]V_{i}^2[/tex]

Look, use your ALGBEBRA skills to solve for [tex] \Delta s [/tex]. Anything else, you're wasting your time.

You are close, you have the [tex]\Delta x[/tex] as some constant * [tex]V_{i}^2[/tex]

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- #24

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this is wat i got then (-750v^2) / (735) =delta s

- #25

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Solve it algebraically, then plug in numbers. You will see that mass does not matter.

[tex]\Delta s =\frac{1}{2\mu g} *v_{i}^2[/tex]

[tex]\Delta s =\frac{1}{2\mu g} *v_{i}^2[/tex]

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- #26

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ok this is what i get

(1/2)MV^2 / "mu"MG = delta s

(1/2)MV^2 / "mu"MG = delta s

- #27

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Right, so the "M"s cancel.

- #28

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so i get (1/2)V^2 / "mu" G = delta s

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