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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 | 0 m/s |
| 3.2 m | 6 m/s |
| 0 m | 10 m/s |
Show by calculations that the above data verifies the law of conservation of energy (Neglect air resistance). (g = 10 m/s2).
Solution
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 = 1/2 mv^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,
`K.E = 1/2 mv^2`
= `1/2 xx (1) xx (0)` J
= 0 J
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,
`K.E = 1/2 mv^2`
= `1/2 xx (1) xx (6)^2` J
= 18 J
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,
`K.E = 1/2 mv^2`
= `1/2 (1)(10)^2` J
= 50 J
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.
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