# I need help solving an equation

• Apr 4th 2010, 09:12 AM
Musturd
I need help solving an equation
I'm trying to do an air resistance lab problem for physics class.
Basically, I had two sets of data (mass and velocity) and plugged them in to this equation: m(9.8) = kv^n

I solved up to a certain point (here):
31.16 = k(2.70)^((20.78 – ln(k))/(.419))

In the end, I need to find n, but I figure that if I can solve for k in that equation then I can easily find n.

My teacher said we can just guess and check for this problem, but I'd rather have a solution for the equation.
(And of course I will credit whoever helps me out)

This is due on Tuesday, so I hope someone can help me out sometime tonight or tomorrow.

Thanks
• Apr 4th 2010, 10:44 AM
skeeter
Quote:

Originally Posted by Musturd
I'm trying to do an air resistance lab problem for physics class.
Basically, I had two sets of data (mass and velocity) and plugged them in to this equation: m(9.8) = kv^n

I solved up to a certain point (here):
31.16 = k(2.70)^((20.78 – ln(k))/(.419))

In the end, I need to find n, but I figure that if I can solve for k in that equation then I can easily find n.

My teacher said we can just guess and check for this problem, but I'd rather have a solution for the equation.
(And of course I will credit whoever helps me out)

This is due on Tuesday, so I hope someone can help me out sometime tonight or tomorrow.

Thanks

$\displaystyle 31.16 = k(2.70)^b$ , where $\displaystyle b = \frac{20.78-\ln{k}}{.419}$

you're not going to be able to solve the given equation using elementary algebraic methods.

solving your equation using a calculator ...

$\displaystyle k \approx 3.308 \times 10^{14}$

which leads me to believe something may be in error.

Correct me if I'm mistaken, but isn't the equation $\displaystyle mg = kv^n$ valid only when the falling mass is at terminal velocity?

do you have some representative data that you can share?
• Apr 4th 2010, 11:55 AM
Musturd
Wow.
Here is my data:
(2.12g, 1.52 m/s)
(3.18g, 2.7 m/s)

I forgot to convert the grams to kilograms -- maybe I didn't screw up my math.

So with the conversion the original equation would be:
$\displaystyle .3112 = k(2.70)^b$ , where $\displaystyle b = \frac{.2078-\ln{k}}{.419}$

Anyway, since this can't be solved with algebra (like you said), I plugged it into wolfram|alpha: http://www.wolframalpha.com/input/?i=(.3312)+%3D+k*(2.70)^((.2078+%E2%80%93+ln(k))/(.419)))

$\displaystyle k \approx 3.208$
looks promising
...
so $\displaystyle n \approx -2.28$
Which is not good. n should be around +2.28 not -2.28
Maybe I did do my beginning steps wrong.

Here are my beginning steps:
(.0212)(9.8) = k*(1.52)n
.2078 = ln(k*1.52n)
.2078 = ln(k) + n*ln(1.52)
(.2078 – ln(k))/(ln(1.52)) = n
Then
$\displaystyle (.0318)(9.8) = k(2.70)^n$ , where $\displaystyle n = \frac{.2078-\ln{k}}{ln{1.52}}$
• Apr 4th 2010, 11:59 AM
Musturd
Wow.
Here is my data:
(2.12g, 1.52 m/s)
(3.18g, 2.7 m/s)

I forgot to convert the grams to kilograms -- maybe I didn't screw up my math.

So with the conversion the original equation would be:
$\displaystyle .3112 = k(2.70)^b$ , where $\displaystyle b = \frac{.2078-\ln{k}}{.419}$

Anyway, since this can't be solved with algebra (like you said), I plugged it into wolfram|alpha: http://www.wolframalpha.com/input/?i=(.3312)+%3D+k*(2.70)^((.2078+%E2%80%93+ln(k))/(.419)))

$\displaystyle k \approx 3.208$
looks promising
...
so $\displaystyle n \approx -2.28$
Which is not good. n should be around +2.28 not -2.28
Maybe I did do my beginning steps wrong.

Here are my beginning steps:
(.0212)(9.8) = k*(1.52)^n
.2078 = ln(k*1.52^n)
.2078 = ln(k) + n*ln(1.52)
(.2078 – ln(k))/(ln(1.52)) = n
Then
$\displaystyle (.0318)(9.8) = k(2.70)^n$ , where $\displaystyle n = \frac{.2078-\ln{k}}{ln{1.52}}$
• Apr 4th 2010, 03:43 PM
skeeter
$\displaystyle mg = kv^n$

setting up a ratio ...

$\displaystyle \frac{.00212g = k(1.52)^n}{.00318g = k(2.7)^n}$

$\displaystyle \frac{.00212}{.00318} = \left(\frac{1.52}{2.7}\right)^n$

now solve for $\displaystyle n$
• Apr 4th 2010, 06:16 PM
Musturd
Thanks for an easy way to calculate it.
I can't believe I still was off by a zero on the kilogram conversion.