Question

The following data was obtained for a body of mass 1 kg dropped from a height of 5 metres :

Distance above ground	Velocity
5 m 3.2 m 0 m	0 m/s 6 m/s 10 m/s

Show by calculations that the above data verifies the law of conservation of energy (Neglect air resistance). (g = 10 m/s²).

Answer 1

Mass of the body, (m₁) = 1 kg
Acceleration due to gravity, (g) = 10 m/s²
We can calculate the potential energy as,
Potential energy = (Weight of body) × (Vertical distance)
Now, we can find the kinetic energy as,
$K.E=\frac{1}{2}m{v}^2$
We have three cases:

Case – 1
Height (h) = 5 m
Velocity (v) = 0 m/s
So,
Potential energy = (Weight of body) × (Vertical distance)
1 × 10 × 5
= 50 J
And,
$$\begin{gathered} K.E=\frac{1}{2}m{v}^2 \\ =\frac{1}{2}\times \left(1\right)\times \left(0\right)\mathrm{J} \\ =0\mathrm{J} \end{gathered}$$
So total energy at this instant,
P.E + K.E
= (50 + 0) J
= 50 J

Case – 2
Height (h) = 3.2 m
Velocity (v) = 6 m/s
So,
Potential energy = (Weight of body) × (Vertical distance)
= (1) × (10) × (3.2) J
= 32 J
And,
$$\begin{gathered} K.E=\frac{1}{2}m{v}^2 \\ =\frac{1}{2}\times \left(1\right)\times (6{)}^2\mathrm{J} \\ =18\mathrm{J} \end{gathered}$$
So total energy at this instant,
P.E + K.E
= (32 + 18) J
= 50 J

Case – 3
Height (h) = 0 m
Velocity (v) = 10 m/s
So,
Potential energy = (Weight of body) × (Vertical distance)
= (1) × (10) × (0) J
= 0 J
And,
$$\begin{gathered} K.E=\frac{1}{2}m{v}^2 \\ =\frac{1}{2}\left(1\right)(10{)}^2\mathrm{J} \\ =50\mathrm{J} \end{gathered}$$
So total energy at this instant,
= P.E + K.E
= (0 + 50) J
= 50 J
We can observe that total energy in all the above cases is same, which satisfies the law of conservation of energy.