A 17-kg rock is on the edge of a 103-m cliff.
(a) What potential energy does the rock possess relative to the base of the cliff?
1 . J
(b) The rock falls from the cliff. What is its kinetic energy just before it strikes the ground?
2 . J
(c) What speed does the rock have as it strikes the ground?
3 44.7m/s
- i got the 3 answer, need help with the other 2 setups!
(a) PE = m•g•h
(b) KE = PE
(c) mv²/2 = PE,
v =sqrt(2•PE/m)
can you show me how to plug each of them in?
To find the potential energy of the rock at the edge of the cliff, we can use the formula:
Potential energy = mass × gravitational acceleration × height
In this case, the mass of the rock is 17 kg, the gravitational acceleration is approximately 9.8 m/s^2, and the height of the cliff is 103 m.
(a) Potential energy of the rock relative to the base of the cliff:
Potential energy = 17 kg × 9.8 m/s^2 × 103 m
Potential energy = 17 × 9.8 × 103
Potential energy ≈ 16,994 J
Therefore, the potential energy of the rock relative to the base of the cliff is approximately 16,994 J.
(b) To find the kinetic energy of the rock just before it strikes the ground, we can use the formula:
Kinetic energy = (1/2) × mass × velocity^2
In this case, the mass of the rock is still 17 kg. We need to find the velocity of the rock just before it strikes the ground.
(c) Since the rock falls from a height, the potential energy it possesses at the edge of the cliff gets converted to kinetic energy just before it strikes the ground.
The conversion of potential energy to kinetic energy can be expressed by the formula:
Potential energy = Kinetic energy
Substituting the values we already know:
16,994 J = (1/2) × 17 kg × velocity^2
Simplifying the equation:
8,497 J = 8.5 kg × velocity^2
To find the velocity, we divide both sides of the equation by 8.5 kg:
velocity^2 = 8,497 J / 8.5 kg
velocity^2 ≈ 1,000
Taking the square root of both sides of the equation, we find:
velocity ≈ √1,000
velocity ≈ 31.6 m/s
Therefore, the speed of the rock as it strikes the ground is approximately 31.6 m/s.
To find the potential energy of the rock at the edge of the cliff, you can use the formula:
Potential Energy = mass x gravitational acceleration x height
Given:
mass (m) = 17 kg
height (h) = 103 m
gravitational acceleration (g) = 9.8 m/s^2 (approx.)
(a) Calculate the potential energy:
Potential Energy = 17 kg x 9.8 m/s^2 x 103 m
Multiply these values together to get the potential energy in joules:
Potential Energy = 17 kg x 9.8 m/s^2 x 103 m = 17 x 9.8 x 103 J = 17,086 J (rounded to nearest whole number)
Therefore, the potential energy of the rock relative to the base of the cliff is approximately 17,086 J.
(b) To find the kinetic energy just before the rock strikes the ground, you can use the formula:
Kinetic Energy = 0.5 x mass x velocity^2
Since the rock is falling freely, we know that the gravitational potential energy it has at the top of the cliff is converted entirely into kinetic energy at the bottom of the cliff. Therefore, the potential energy at the top equals the kinetic energy at the bottom.
The potential energy at the top is 17,086 J (as calculated in part a), so:
Kinetic Energy = 17,086 J
Therefore, the kinetic energy of the rock just before it strikes the ground is also approximately 17,086 J.
(c) The speed at which the rock strikes the ground can be found by using the kinetic energy equation:
Kinetic Energy = 0.5 x mass x velocity^2
Given:
Kinetic Energy = 17,086 J (as calculated in part b)
mass (m) = 17 kg
Let's rearrange the equation to solve for velocity (v):
17,086 J = 0.5 x 17 kg x v^2
Divide both sides of the equation by 0.5 x 17 kg:
v^2 = (2 x 17,086 J) / 17 kg
v^2 = 338,960 / 17 kg
v^2 = 19,940 m^2/s^2
To find the speed (v), take the square root of both sides:
v = √(19,940 m^2/s^2)
v ≈ 141.4 m/s (rounded to one decimal place)
Therefore, the speed at which the rock strikes the ground is approximately 141.4 m/s.