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A 0.10 kg stone is thrown from the edge of an ocean cliff with an initial speed
of 20 m/s. When it strikes the water below, it is traveling at 45 m/s. What is the
change in the kinetic energy of the stone?

Sagot :

Answer:

79.20 joule

Explanation:

1.

A 0.10kg stone is thrown from the edge of an ocean cliff with an initial speed of 21m/s. When it strikes the water below, it is traveling at 45m/s. What is the change in kinetic energy of the stone?

ΔEk = m/2*(V2^2-V1^2) = 0.05*(45^2-21^2) = 79.20 joule

The change in the kinetic energy of the stone is 81.25 J.

The given parameters;

  • mass of the stone, m = 0.1 kg
  • initial speed of stone, u = 20 m/s
  • final velocity of the stone, v = 45 m/s

The initial kinetic energy of the stone is calculated as follows;

K.E = ¹/₂mu²

K.E₁ = ¹/₂ x 0.1 x 20²

K.E₁ = 20 J

The final kinetic energy of the stone is calculated as follows;

K.E₂ = ¹/₂ x 0.1 x 45²

K.E₂ = 101.25 J

The change in the kinetic energy of the stone is calculated as follows;

ΔK.E = K.E₂ - K.E₁

ΔK.E = 101.25 J  -  20 J

ΔK.E = 81.25 J

Thus, the change in the kinetic energy of the stone is 81.25 J.

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