Find the increase in kinetic energy of a body of mass 500 g ,when its speed
increases from 2 m/s to 4 m/s.
Answers
given,
initial velocity, u = 2 m/s( v 1)
final velocity, v = 4m/s ( v 2)
mass of the body = 500 g
= 500/1000 kg
= 1/2 kg
Increase in kinetic energy = 1/2 m × ((v2) ^2 - (v1) ^2)
= 1/2 × 1/2 × (( 4)^2 - (2)^2
= 1/4 × ( 16 - 4)
= 1/4 × 12
= 3 joule
Answer :
The increase of kinetic energy of the body is 3 Joule.
Explanation :
Given that,
- Mass of body = 0.5kg [ 500g = 0.5kg
- Increase of speed is 2m/s to 4m/s
Let us assume the increase in Kinetic energy of the body as K.E.₂ and the another as K.E.₁
First we need to find out the Kinetic energies of K.E.₁ and K.E.₂
As we know that,
K.E. = {1/2} × m × v²
Where,
- m = Mass of the body.
- v = Speed of the body.
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The kinetic energy of K.E.₁
➠ 1/2 × 0.5 × 2²
➠ 1/2 × 0.5 × 2 × 2
➠ 0.5 × 2
➠ 1
The kinetic energy of K.E.₁ is 1 Joule.
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The kinetic energy of K.E.₂
➠ 1/2 × 0.5 × 4²
➠ 1/2 × 0.5 × 4 × 4
➠ 0.5 × 2 × 4
➠ 1 × 4
➠ 4
The kinetic energy of K.E.₂ is 4 Joule.
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Therefore, The increase of kinetic energy of the body is,
➠ K.E.₂ - K.E.₁
➠ 4J - 1J
➠ 3J ★
Hence, the increase of kinetic energy of the body is 3 Joule.