A 17 kg rock, m, is on the edge of a 105 m cliff, h, as shown in Figure 11-20. Assume that air resistance is negligible.
What speed does the rock have as it strikes the ground?
Well, that rock sure seems to have a lot on its mind, like falling off cliffs and calculating its speed! But don't worry, I'll be here to give you a laugh and answer your question.
Now, let's get back to the rock's speed. We can use the law of conservation of energy to help us out here. When the rock is at the edge of the cliff, it has potential energy due to its height, which can be converted into kinetic energy as it falls.
The potential energy of the rock at the edge of the cliff can be calculated using the formula: PE = mgh, where m is the mass of the rock (17 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the cliff (105 m).
So, the potential energy of the rock is: PE = (17 kg) * (9.8 m/s^2) * (105 m).
Now, when the rock strikes the ground, all of its potential energy is converted into kinetic energy. The kinetic energy of an object is given by the formula: KE = (1/2)mv^2, where m is the mass of the rock and v is its speed.
Equating the potential energy to the kinetic energy, we get: PE = KE.
(17 kg) * (9.8 m/s^2) * (105 m) = (1/2) * (17 kg) * v^2.
Now, we can solve for v! I'll whip out my calculator to crunch the numbers… *beep boop beep*
Huh, looks like the speed the rock has as it strikes the ground is approximately 45 m/s. But hey, just remember that even a falling rock can't resist the force of gravity and eventually has to face the ground!
To find the speed at which the rock strikes the ground, we can use the principle of conservation of energy. At the edge of the cliff, the rock has gravitational potential energy, which is then converted into kinetic energy as it falls.
1. Calculate the gravitational potential energy of the rock at the edge of the cliff using the formula:
Potential Energy = mass * gravity * height
where the mass of the rock (m) is 17 kg, gravity (g) is 9.8 m/s^2, and the height of the cliff (h) is 105 m.
Potential Energy = 17 kg * 9.8 m/s^2 * 105 m = 17865 J (joules)
2. This potential energy is then converted into kinetic energy. The formula for kinetic energy is:
Kinetic Energy = (1/2) * mass * velocity^2
Using the equation for potential energy from step 1, we can set it equal to the kinetic energy:
17865 J = (1/2) * 17 kg * velocity^2
Simplify the equation:
35730 J = 17 kg * velocity^2
Divide both sides of the equation by 17 kg:
2101 J/kg = velocity^2
3. Take the square root of both sides of the equation to solve for the velocity:
velocity = √(2101 J/kg)
velocity ≈ 45.89 m/s
Therefore, the speed at which the rock strikes the ground is approximately 45.89 m/s.
To find the speed at which the rock strikes the ground, we can use the concept of conservation of energy. The gravitational potential energy at the top of the cliff is converted into kinetic energy when the rock strikes the ground. Therefore, we can equate the initial potential energy at the top of the cliff to the final kinetic energy at the bottom.
The potential energy of an object at height h can be calculated using the formula:
Potential energy = mgh
where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
The initial potential energy of the rock at the top of the cliff is given by:
Initial potential energy = mgh
Substituting the given values, we have:
Initial potential energy = 17 kg * 9.8 m/s^2 * 105 m
Next, we equate the initial potential energy to the final kinetic energy at the bottom of the cliff.
The kinetic energy of an object can be calculated using the formula:
Kinetic energy = (1/2) * mv^2
where m is the mass and v is the velocity.
At the bottom of the cliff, the rock will have only kinetic energy, so the final kinetic energy is given by:
Final kinetic energy = (1/2) * mv^2
We can equate the initial potential energy to the final kinetic energy:
17 kg * 9.8 m/s^2 * 105 m = (1/2) * 17 kg * v^2
Now, we can solve for the velocity (v):
v^2 = (2 * 17 kg * 9.8 m/s^2 * 105 m) / 17 kg
v^2 = 2 * 9.8 m/s^2 * 105 m
v^2 = 2058 m^2/s^2
To find the velocity (v), we take the square root of both sides:
v = √(2058 m^2/s^2)
v ≈ 45.4 m/s
Therefore, the speed at which the rock strikes the ground is approximately 45.4 m/s.