Answer:
There will be an increase in the kinetic energy
Explanation:
A falling object converts the gravitational potential energy to the kinetic energy. The potential energy is then converted to kinetic energy followed by the conversation:
[tex]E_{p} = E_{k}[/tex]
where Ep and Ek are potential and kinetic energies respectively.
This potential energy is then converted to kinetic energy. Halfway, the kinetic energy is equal to KE1.
However, the kinetic energy is given by the equation:
[tex]KE = \frac{1}{2}mv^{2}[/tex]
As the velocity increases, the kinetic energy increases. Hence KE2 will be greater than KE1
When K.E₂ is compared to twice K.E₁, the K.E₂ increases by factor of 8 when K.E₁ is doubled.
The given parameters;
The kinetic energy of the object is calculated as follows;
[tex]K.E = \frac{1}{2} mv^2[/tex]
Based on the principle of conservation of energy;
K.E = P.E
When the object has fallen halfway;
[tex]K.E = \frac{mgh}{2} \\\\K.E_1 = \frac{mgh}{2}[/tex]
When the object has fallen twice as far;
[tex]K.E = mg(2h)\\\\K.E_2 = 2mgh\\\\K.E_2 = 2(mgh)\\\\K,E_2 = 2\times \frac{2}{2} \times mgh\\\\K.E_2 = 4 \times \frac{1}{2} \times mgh\\\\K.E_2 = 4 \times \frac{mgh}{2} \\\\K.E_2 = 4(K.E_1)[/tex]
To compare K.E₂ to twice K.E₁;
[tex]K.E_2 = 4(K.E_1)\\\\K.E_2 = 4(2K.E_1)\\\\K.E_2 = 8K.E_1[/tex]
Thus, we can conclude that when K.E₂ is compared to twice K.E₁, the K.E₂ increases by factor of 8 when K.E₁ is doubled.
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