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Sagot :
Answer:
968 J
Explanation:
The computation of the kinetic energy is shown below:
Given that
mass m = 2 kg
height h = 20 m
velocity v = 24 m / s
Now
According to the law of conservation of energy ,
= K.E at top + P.E at top
= ( 1 ÷ 2) mv ^ 2 + mgh
= 576 + 392
= 968 J
The kinetic energy of the mass just before it strikes the ground is 968 J.
The given parameters:
- Mass of the object, m = 2.0 kg
- Height of the building, h = 20 m
- Initial velocity of the mass, u = 24 m/s
The kinetic energy of the mass just before it strikes the ground is calculated by applying the principle of conservation of energy;
[tex]K.E_f = K.E_ i + P.E_i\\\\K.E _f = \frac{1}{2} mv^2 + mgh\\\\K.E_f = \frac{1}{2} \times 2\times 24^2 \ + \ 2 \times 9.8 \times 20\\\\K.E_f = 968 \ J[/tex]
Thus, the kinetic energy of the mass just before it strikes the ground is 968 J.
Learn more about conservation of energy here: https://brainly.com/question/166559
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