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The following data were obtained for a body of mass 1 kg dropped from a height of 5 meters:
Distance above ground | Velocity |
$5\ m$ | $0\ m/s$ |
$3.2\ m$ | $6\ m/s$ |
$0\ m$ | $10\ m/s$ |
Show by calculations that the above data verify the law of conservation of energy (Neglect air resistance).$(g=10\ m/s^2)$."
The given data of the body can be illustrated as shown in the figure below:
As given, Mass of the body $m=1\ kg$
Gravitational acceleration $g=10\ m/s^2$
At point A $(height=5\ m)$:
Velocity $v=0$ [as given in the data table)
Height $h=5\ m$
So, kinetic energy $K_A=\frac{1}{2}mv^2$
$=\frac{1}{2}\times1\ kg\times 0$
$=0$
Its potential energy $P_A=mgh$
$=1\ kg\times 10\ m/s^2\times 5\ m$
$=50\ J$
Total energy $E=K_A+P_A$
$=0+50\ J$
$=50\ J$
At Point B $(height=3.2\ m)$:
Velocity $v=6\ m/s$
Height $h=3.2\ m$
Therefore, kinetic energy $K_B=\frac{1}{2}mv^2$
$=\frac{1}{2}\times1\ kg\times(6\ m/s)^2$
$=18\ J$
Its Potential energy at point B, $P_B=mgh$
$=1\ kg\times10\ m/s^2\times3.2\ m$
$=32\ J$
Total energy at Point B, $E_B=K_B+P_B$
$=18\ J+32\ J$
$=50\ J$
At point C $(On\ ground,\ h=0)$
Velocity $v_C=10\ m/s$
Height $h=0$
Therefore, kinetic energy $K_C=\frac{1}{2}mv^2$
$=\frac{1}{2}\times 1\ kg\times 10^2$
$=50\ J$
Potential energy $P_C=mgh$
$=1\ kg\times10\ m/s^2\times 0$
$=0$
So, the total energy at the ground$(on\ C)$, $E_C=K_C+P_C$
$=50\ J+0$
$=50\ J$
Here it is clearly observed that total energy remains the same at every point which proves the law of energy conservation.