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2 years ago

A 30.0 kg rock falls from a 35.0 m cliff. What is the kinetic and potential energy when the rock has fallen 12.0 m?

Physics

1 answer:

Arturiano [62]2 years ago

6 0

Answer:

When the rock has fallen 12.0 m;

The kinetic energy of the rock is approximately 3,531.6 J

The potential energy of the rock is approximately 6,768.9 J

Explanation:

The question relates to the characteristic constant total mechanical energy of a body

The mass of the rock that falls, m = 30.0 kg

The height of the cliff from which the rock falls, h₁ = 35.0 m

The required information = The kinetic and potential energy when the rock has falling 12.0 m

The kinetic energy is given by the formula, K.E. = 1/2×m×v²

The potential energy is given by the formula, P.E. = m·g·h

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

The velocity of the rock after falling through a given height, h is given by the formula, v² = 2·g·Δh

The total mechanical energy of the rock, M.E. = K.E. + P.E. = Constant

At the height of the cliff before falling, Δh =0, therefore v₁ = 0, therefore, K.E. = 1/2×m×v₁² = 0 J

The potential energy at the cliff before the rock begins to fall, P.E. is goven as follows;

P.E. = 30.0 kg × 35.0 m × 9.81 m/s² = 10,300.5 J

At the top of the cliff, M.E. = K.E. + P.E. = 0 J+ 10,300.5 J = 10,300.5 J

∴ M.E. = 10,300.5 J

When the rock has fallen, 12.0 m, Δh = 12.0 m, the speed of the rock, v₂, is given as follows;

v₂² = 2 × 9.81 m/s² × 12.0 m = 235.44 m²/s²

v₂ = √(235.44 m²/s²) ≈ 15.344 m/s

∴ When the rock has fallen 12.0 m, K.E., is given as follows;

K.E. = 1/2×m×v₂²

K.E. = 1/2 × 30.0 kg × 235.44 m²/s² = 3,531.6 J

When the rock has fallen 12.0 m the kinetic energy, K.E. = 3,531.6 J

When the rock has fallen 12.0 m, M.E. = P.E. + K.E.

M.E. = Constant = 10,300.5 J

K.E. = 3,531.6 J

∴ 10,300.5 J = P.E. + 3,531.6 J

P.E. = 10,300.5 J - 3,531.6 J = 6,768.9 J

∴ When the rock has fallen 12.0 m, the potential energy, P.E. = 6,768.9 J.

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