På a) fikk vi svaret $x = \frac{2 \sqrt{2}}{3} \cdot v_0 \cdot \sqrt{\frac{m}{k}}$, mens fasiten fikk $x_f = \frac{2}{3} \sqrt{\frac{2m}{k}} v_0$. Så den er jo grei,
5. A mass $m$ on a horizontal surface is attached to a massless, ideal spring (spring constant $k$).
First, assume a frictionless surface:
a)
If the mass is given an initial velocity $v_0$ at $x=0$, at what rightward displacement $x$ has the mass's speed $v$ decreased to $\frac{1}{3}v_0$?
Now, suppose that the surface does have some kinetic friction:
b)
Suppose the mass is again given an initial velocity of $v_0$ at $x=0$. The mass reaches a maximum rightward displacement $d$ before (momentarily) stopping and turning around. Find an expression for the fraction of the system's initial mechanical energy that is lost to friction during the mass's rightward motion.
Show your work. Express your final answer...
* ONLY in terms of variables $m$, $k$, $v_0$, $d$, and mathematical constants
* in SIMPLEST algebraic fom
* with mathematical constants expressed EITHER as simplified pure rational numbers OR as decimal values with three significant figures
Men på b) har vi feil. Fasit sier følgende:
Mens vi gjorde dette:Initial energy (at $x=0$):
$E_{\textrm{init}} = \frac{1}{2}mv^2$
Final energy (at $x=d$):
$E_{\textrm{final}} = \frac{1}{2}kd^2$
Fraction of initial energy lost:
$\frac{\Delta E}{E_{\textrm{init}}} = \frac{E_f - E_i}{E_i} = \frac{\frac{1}{2}kd^2 - \frac{1}{2}mv_0^2}{\frac{1}{2}mv_0^2} = \frac{kd^2 - mv_0^2}{mv_0^2}$
Jeg henger selvfølgelig med på fasit, men jeg klarer ikke å se hvor vi har gjort feil selv. Noen som kan hjelpe?Start:
$E_{\textrm{total}} = E_k = \frac{1}{2}mv^2$
Slutt:
$E_{\textrm{total}} = E_p = \frac{1}{2}kd^2$
Det vil si at på slutten sitter vi igjen med en andel som tilsvarer $\frac{E_p}{E_k}$.
Det betyr at vi har mistet $1 - \frac{E_p}{E_k}$.
Det vi har mistet delt på den opprinnelige energien:
$\frac{1 - \frac{E_p}{E_k}}{E_k}$
Enkel algebra, så får vi: $\frac{\frac{E_k - E_p}{E_k}}{E_k} = \frac{E_k - E_p}{E_k^2}$
Setter inn for $E_p$ og $E_k$:
$\frac{E_k - E_p}{E_k^2} = \frac{\frac{1}{2}mv_0^2 - \frac{1}{2}kd^2}{\frac{1}{4}m^2v_0^4} = \underline{\underline{\frac{2\left(mv_0^2 - kd^2\right)}{m^2v_0^4}}}$
