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A 2.0-kg mass is projected from the edge of the top of a 20-m tall building with a velocity of 24 m/s at some unknown angle above the horizontal. Disregard air resistance and assume the ground is level. What is the kinetic energy of the mass just before it strikes the ground?

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

onyebuchinnaji

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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