Using Energy to find the speed

  • Thread starter swannyboy
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  • #26
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I get something similar, so you might want to check:

v2 = (1/M)*(-k*x2 + 2*(m1*g - Ff)*x
where: M = m1 + m2

Which, when values are plugged in, boils down to

v2 = -33.33(N/kg/m)*x2 + 0.1380(N/kg)*x

I think it must be an equivalent form.

I got that it was the square root of 1/3. I've had this confirmed by 2 people now.
Edit: Nope, you're right. Using exact values, it still works out to be the square root of 1/3.

Well... I tried it using the form you got. I got slightly more than the square root of 1/3. However, I just saw the same question done using de Moivre's theorem, and the answer was the square root 1/3.
 
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  • #27
gneill
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I think it must be an equivalent form.

I got that it was the square root of 1/3. I've had this confirmed by 2 people now.
Edit: Nope, you're right. Using exact values, it still works out to be the square root of 1/3.

Well... I tried it using the form you got. I got slightly more than the square root of 1/3. However, I just saw the same question done using de Moivre's theorem, and the answer was the square root 1/3.

The reason I have my doubts is that your formula for V^2 is negative near x = 0. I would expect V^2 to be positive and increasing.

So, in all its glory,

[tex]m1\;g\;x = \frac{1}{2}(m1 + m2)v^2 + \frac{1}{2}k\;x^2 + F_f\;x[/tex]

I don't see how you could arrive at an exact answer of sqrt(1/3), with the constant g mixed up in the result...

[tex]v^2 = \frac{1}{m1 + m2}(-k\;x^2 + 2\;(m1\;g - F_f)\;x)[/tex]

differentiating, setting to zero, and solving for x yields

[tex]x = \frac{m1\;g - F_f}{k}[/tex]

[tex]x = 2.066 x 10^{-3}\;m[/tex]

It also yields (symbolically)

[tex]v^2_{max} = \frac{(m1\;g - F_f)^2}{k(m1 + m2)}[/tex]

[tex]v = \frac{m1\;g - F_f}{\sqrt{k(m1 + m2)}}[/tex]

[tex]v = 11.93 mm/sec[/tex]
 

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