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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?

Short Answer

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

Step by step solution

01

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.
02

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 \( g = 9.8 \ \text{m/s}^2 \):\[ v_f = \sqrt{2 \times 9.8 \ \text{m/s}^2 \times 95.0 \ \text{m}} = \sqrt{1862} \text{ m/s} \approx 43.2 \ \text{m/s} \].
03

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.
04

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 \( g = 9.8 \ \text{m/s}^2 \):\[ v_i = \sqrt{2 \times 9.8 \ \text{m/s}^2 \times 525 \ \text{m}} = \sqrt{10290} \text{ m/s} \approx 101.4 \ \text{m/s} \].
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

To stretch a spring 3.00 cm from its unstretched length, 12.0 J of work must be done. (a) What is the force constant of this spring? (b) What magnitude force is needed to stretch the spring 3.00 cm from its unstretched length? (c) How much work must be done to compress this spring 4.00 cm from its unstretched length, and what force is needed to compress it this distance?

A mass \(m\) slides down a smooth inclined plane from an initial vertical height \(h\), making an angle \(\alpha\) with the horizontal. (a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height \(h\). (b) Use the work\(-\)energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if the mass had been dropped from height \(h\), independent of the angle \(\alpha\) of the incline. Explain how this speed can be independent of the slope angle. (c) Use the results of part (b) to find the speed of a rock that slides down an icy frictionless hill, starting from rest 15.0 m above the bottom.

A spring of force constant 300.0 N/m and unstretched length 0.240 m is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to 15.0 N. How long will the spring now be, and how much work was required to stretch it that distance?

You apply a constant force \(\overrightarrow{F} =(-68.0 \, \mathrm{N})\hat{\imath} +(36.0 \, \mathrm{N})\hat{\jmath}\) to a 380-kg car as the car travels 48.0 m in a direction that is 240.0\(^\circ\) counterclockwise from the +\(x\)-axis. How much work does the force you apply do on the car?

Use the work\(-\)energy theorem to solve each of these problems. You can use Newton's laws to check your answers. (a) A skier moving at 5.00 m/s encounters a long, rough horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. How far does she travel on this patch before stopping? (b) Suppose the rough patch in part (a) was only 2.90 m long. How fast would the skier be moving when she reached the end of the patch? (c) At the base of a frictionless icy hill that rises at 25.0\(^\circ\) above the horizontal, a toboggan has a speed of 12.0 m/s toward the hill. How high vertically above the base will it go before stopping?

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