Mechanics11/Seite11/Loesung4.7

Angabe

 * Masse m = 70g = 0,070 kg
 * Geschwindigkeit v0 = 5,0 m/s
 * Höhe h = 1,0 m

Rechnung
$$\;a)$$ $$E_{kin} = \frac{1}{2}*m*v^2$$

$$E_{kin} = \frac{1}{2}*0,070*5,0^2 J$$

$$E_{kin} = \frac{}{} 0,875 J$$

$$E_{pot} = \frac{}{} m*g*h$$

$$E_{pot} = \frac{}{} m*g*2r$$

$$E_{pot} = \frac{}{} 0,070*9,81*2*0,50 J$$

$$E_{pot} = \frac{}{} 0,687 J$$

$$\;b)$$

$$\frac{}{}E_{kin'} = E_{kin0}$$

$$\frac{}{} => v' = v_0$$

--FatF 09:12, 22 February 2008 (UTC)