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.
