# $\mu_k$ of surface from experimental vs theoretical acceleration [closed]

I have an air track at a slight angle with a glider. Final velocity is determined, and from that a is calculated, and from a, I have approximated g. I am trying to find $$\mu_{k}$$ between the track and the glider. The values that I am coming up with are about 1/100th of what I expect. What I have so far is:

$$g=\frac{a_{exp}}{\sin\theta}$$

$$a_{exp}=\frac{v_{f}^{2}}{2d}$$

$$g=\frac{\frac{v_{f}^{2}}{2d}}{\sin\theta}$$

$$v_{t_{f}}=\sqrt{g\left(2d\right)\sin\theta}$$

$$K_{exp}=\frac{1}{2}mv^{2}_{exp_{f}}$$

$$K_{t}=\frac{1}{2}mv^{2}_{t_{f}}$$

$$K_{loss}=K_{t}-K_{exp}$$

$$F_{fric}=\frac{K_{loss}}{d}$$

$$F_{n}=mg\cos\theta$$

$$\mu_{k}=\frac{F_{fric}}{F_{n}}$$

$$\mu_{k}=\frac{\frac{\frac{1}{2}m\sqrt{g\left(2d\right)}^{2}-\frac{1}{2}mv^{2}_{exp_{f}}}{d}}{mg\cos\theta}$$

Am i missing something obvious, or am I way off the mark?

## closed as off-topic by John Rennie, Aaron Stevens, Jon Custer, ZeroTheHero, tpg2114♦Aug 16 at 23:17

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• You need to give more information. All you have given is a derivation of an expression. What makes you think this expression is wrong? What did you measure? Where are you getting your expected value from? – Aaron Stevens Aug 14 at 10:40
• @AaronStevens the expected value for mu k was given by the professor to be in or around the second decimal place. I feel like my reasoning is correct, but I am getting values between 0.0018 and 0.0004 with a mean of 0.0008. I expect it to be negligible as it's and air track, just not this negligible. the measurements taken in the original experiment were of a glider along an inclined track across variable distances from an initial velocity of 0 through a photogate timer to get final velocity at end of run. from this g was calculated by dividing by the sine of the angle. – theoctagon Aug 14 at 16:18