Problem 21 - University Physics with Modern Physics

Question

Use the work\(-\)energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases. (a) A branch falls from the top of a 95.0-m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 m into the air. How fast was the boulder moving just as it left the volcano?

Step-by-Step Solution

Answer

(a) 43.2 m/s when it reaches the ground. (b) Boulder initial speed is 101.4 m/s.

Step 1: Understand the Work-Energy Theorem

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. Mathematically, it can be expressed as \( W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 \). When air resistance is neglected, gravitational potential energy can convert fully to kinetic energy when an object falls or rises.

Step 2: Part a: Calculate the Speed of a Falling Branch

Using the work-energy principle, the initial potential energy \( PE_i = mgh \) will convert entirely into kinetic energy \( KE_f = \frac{1}{2}mv_f^2 \) when the branch reaches the ground. As it falls from rest, \( v_i = 0 \). So, \( mgh = \frac{1}{2}mv_f^2 \) simplifies to \( gh = \frac{1}{2}v_f^2 \). Solve for \( v_f \):1. \( v_f^2 = 2gh \)2. \( v_f = \sqrt{2gh} \)Plug in \( h = 95.0 \) m and \( g = 9.8 \ \text{m/s}^2 \):\[ v_f = \sqrt{2 \times 9.8 \ \text{m/s}^2 \times 95.0 \ \text{m}} = \sqrt{1862} \text{ m/s} \approx 43.2 \ \text{m/s} \].

Step 3: Verify Part a with Newton's Laws

Using kinematic equations: \( v_f^2 = v_i^2 + 2gh \). Here, \( v_i = 0 \), and substituting known values:\( v_f^2 = 2 \times 9.8 \ \text{m/s}^2 \times 95.0 \ \text{m} \rightarrow v_f = \sqrt{1862} \text{ m/s} \approx 43.2 \ \text{m/s} \). The result confirms the previous calculation.

Step 4: Part b: Calculate Initial Speed of Ejected Boulder

The boulder was provided an initial kinetic energy corresponding to its potential height of 525 m. Using \( KE_i = PE_f \), the formula for the potential energy (at maximum height) \( mgh = \frac{1}{2}mv_i^2 \) simplifies to:1. \( gh = \frac{1}{2}v_i^2 \)2. \( v_i^2 = 2gh \)3. \( v_i = \sqrt{2gh} \)Substitute \( h = 525 \) m and \( g = 9.8 \ \text{m/s}^2 \):\[ v_i = \sqrt{2 \times 9.8 \ \text{m/s}^2 \times 525 \ \text{m}} = \sqrt{10290} \text{ m/s} \approx 101.4 \ \text{m/s} \].

Step 5: Verify Part b with Newton's Laws

Using kinematic equations: \( v_i^2 = v_f^2 + 2gh \). Since the boulder reaches maximum height at \( v_f = 0 \),\( v_i^2 = 2 \times 9.8 \ \text{m/s}^2 \times 525 \ \text{m} \rightarrow v_i = \sqrt{10290} \text{ m/s} \approx 101.4 \ \text{m/s} \). This calculation confirms the result.

Key Concepts

Kinetic Energy

Kinetic energy is a type of energy that a body possesses due to its motion. It plays a crucial role in the work-energy theorem, which we used in solving our exercises. The kinetic energy of an object with mass \( m \) and moving at velocity \( v \) is given by the formula:

• \( KE = \frac{1}{2}mv^2 \)

The significance of kinetic energy lies in its ability to quantify how much energy is involved when an object changes speed. For example, when a branch falls from a tree, its gravitational potential energy transforms into kinetic energy, leading the branch to accelerate as it descends. This energy transformation follows the principle of energy conservation, assuming no energy is lost to air resistance or friction.

In the exercise with the boulder ejected by the volcano, the initial kinetic energy propels the boulder upwards, and as it rises, this energy converts to potential energy. Understanding the relationship between these energy forms helps in predicting the body's speed at different points in its motion.

Potential Energy

Potential energy is the energy stored within an object due to its position in a force field, such as gravity. It's a vital concept when analyzing systems where energy transformation occurs. The potential energy of an object of mass \( m \) raised to a height \( h \) in a gravitational field is represented by:

• \( PE = mgh \)

Consider the falling branch from a 95-meter-tall tree. Its initial state is stationary with its energy stored as gravitational potential energy. As it falls, this energy is converted into kinetic energy, causing the branch to accelerate until it impacts the ground, consistent with the work-energy theorem.

In the case of the volcano's boulder, the boulder at its peak height holds maximum potential energy and no kinetic energy if we disregard air resistance. This energy conversion allows the calculation of initial speeds necessary to reach certain heights, emphasizing the practical utility of potential energy in solving physics problems.

Newton's Laws

Newton's laws of motion provide a foundational framework for understanding the movement of objects. They are vital for examining how forces affect the motion, and they overlap with energy principles significantly. Here's a brief summary of Newton's three laws:

• First Law: An object will remain at rest or in uniform motion unless acted upon by a net external force.
• Second Law: The acceleration of an object depends on the net force acting on it and is inversely proportional to its mass \( F = ma \).
• Third Law: For every action, there is an equal and opposite reaction.

Newton's laws help verify results in the work-energy theorem calculations. For instance, in the branch falling scenario, we can apply the second law. By equating gravitational force to mass times acceleration, and plugging in the known values, the derived speed from energy considerations matches with the speed calculated using Newton's laws as checked.

Similarly, in the volcano boulder scenario, Newton's laws provide a backdrop to understanding the forces involved, and by calculating acceleration due to gravity, initial velocity can be confirmed. Thus, Newton's laws serve not only to verify but also to provide insight into dynamic systems analyzed through energy principles.