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Sagot :
Answer:
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.
Explanation:
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
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
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.
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.
Calculation of the potential energy and kinetic energy:
Since
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
We know that
P.E = mgh
P.E = (2)(9.8)(20)
P.E= 392 J
Now
K.E = ¹/₂mv²
here.
v is velocity of the rock = 0
K.E = ¹/₂(2)(0)²
K.E = 0
Now
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
Here,
v is the velocity of the rock= 14 m/s
K.E = ¹/₂(2)(14)²
K.E = 196 J.
Learn more about energy here: https://brainly.com/question/13203990
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