Problem 105 - Physics for Scientists and Engineers with Modern Physics

Question

(a) If a volcano spews a \(450-\mathrm{kg}\) rock vertically upward a distance of \(320 \mathrm{m},\) what was its velocity when it left the volcano? \((b)\) If the volcano spews 1000 rocks of this size every minute, estimate its power output.

Step-by-Step Solution

Answer

(a) 79.19 m/s. (b) 23.52 MW.

Step 1: Identify Known Values and Required Formula

We know the mass of the rock, \( m = 450 \) kg, the height it reaches, \( h = 320 \) m, and the acceleration due to gravity \( g = 9.8 \) m/s². We need to find the initial velocity \( v \) using the relation between potential energy and kinetic energy: \( \frac{1}{2}mv^2 = mgh \).

Step 2: Derive Initial Velocity

Rearrange the energy equation to find \( v \):\[ \frac{1}{2}v^2 = gh \] and \[ v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 320} \]. Compute \( v \) to get \( v \approx 79.19 \) m/s.

Step 3: Compute Power Output

First, calculate the energy required for one rock: \( E = mgh = 450 \times 9.8 \times 320 \). Calculate \( E \) to get \( E = 1,411,200 \) J. Since 1000 rocks are ejected every minute, total energy per minute is \( 1,411,200,000 \) J.

Step 4: Convert Energy to Power

Power is energy per time. Convert time from minutes to seconds: 1 minute = 60 seconds. Thus the power \( P = \frac{1,411,200,000}{60} \approx 23,520,000 \) W or 23.52 MW.

Key Concepts

Kinetic energy

Kinetic energy is the energy which an object possesses due to its motion. When we think about a rock being launched from a volcano, it's important to remember that initially, all its energy comes in the form of kinetic energy. This energy is expressed in the formula:

• \[ KE = \frac{1}{2}mv^2 \]

The formula uses the mass \( m \) and the velocity \( v \) to calculate kinetic energy. In our volcano problem, this energy is what propels the rock upward.
The kinetic energy helps transform the rock from mere volcanic mass to a projectile moving skyward with a specific velocity, in this case, around 79.19 m/s when it leaves the volcano.
Understanding this transformation is crucial because it helps explain how energy is conserved and converted from one form to another, such as from kinetic to potential energy as the rock ascends.

Potential energy

Potential energy in this context is the energy stored by the rock due to its position relative to the Earth. This is significant when considering how high the rock rises in the air.
Potential energy can be calculated when the rock reaches its maximum height using the formula:

• \[ PE = mgh \]

where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is the height.
In the volcano problem, as the rock rises, its velocity decreases until all kinetic energy is converted to potential energy at the peak. At the top, the entire kinetic energy has transformed into potential energy, meaning the rock has stored energy equal to the height it has reached — 320 m in this case.
This conversion shows how energy shifts between forms but remains constant in total, demonstrating the law of energy conservation.

Power output

Power output measures how quickly energy is being used or transferred. When discussing the volcanic eruption, assessing the power output gives us an idea of the energy clocking every minute by the exploding volcano, ejecting rocks.
The formula for power is quite straightforward:

• \[ P = \frac{E}{t} \]

where \( E \) is energy and \( t \) is time.
The problem states that the volcano is hurling a considerable mass of rocks (1000 rocks) every minute.
To find the power output of the volcano, we first calculate the potential energy for one rock and then amplify that by the number of rocks ejected per unit of time. Converting the time from minutes to seconds offers a more practical expression of power, giving 23.52 MW. This immense power outlines the intensity and scale at which geological processes like volcanicism can occur.

Velocity calculation

Calculating the velocity of an object launched vertically involves understanding the relationship between different energy forms.
Since kinetic energy is initially used to propel the rock, it is converted into potential energy as the rock ascends, reaching maximum at the pinnacle of its flight.
To calculate this initial velocity, we use the relation between kinetic and potential energy, simplified to:

• \[ v = \sqrt{2gh} \]

where \( g \) is gravity and \( h \) is the height. By substituting the known values, we achieve our velocity.
For this example, solving \( v = \sqrt{2 \times 9.8 \times 320} \), provides approximately \( v = 79.19 \) m/s.
This method shows how energy considerations provide insight into velocity, offering a pathway to solving many real-world motion and force-related problems.