Question

A 2kg rock is at the edge of a cliff 20m above the surface of a lake. It comes loose and falls. Half

way down it is falling at 14 m/s. What is its Kinetic Energy at the moment it hits the water?

Please show work!

Answer 1

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Answer:
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• The potential energy (P.E) at the top is 392 J.
• The kinetic energy (K.E) at the top is 0 J.
• The potential energy (P.E) at the halfway point is 196 J.
• The kinetic energy (K.E) at the halfway point is 196 J.

Step-by-step explanation:

Facts - Key Details Given:

• mass of the rock: m = 2 kg
• height of the cliff: h = 20 m
• speed of the rock at the halfway point: v = 14 m/s
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Calculations:

#1: The potential energy (P.E) and kinetic energy (K.E) when its at the top;

P.E = mgh

P.E = (2)(9.8)(20)

P.E= 392 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the top of the cliff = 0

K.E = ¹/₂(2)(0)²

K.E = 0

#2: The potential energy (P.E) and kinetic energy (K.E) at the halfway point;

P.E = mg(¹/₂h)

P.E = (2)(9.8)(¹/₂ x 20)

P.E = 196 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the halfway point = 14 m/s

K.E = ¹/₂(2)(14)²

K.E = 196 J.