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.

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Updated on: 10-Oct-2022

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