/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 You throw a 20 -N rock verticall... [FREE SOLUTION] | 91Ó°ÊÓ

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You throw a 20 -N rock vertically into the air from ground level. You observe that when it is 15.0 \(\mathrm{m}\) above the ground, it is trav- eling at 25.0 \(\mathrm{m} / \mathrm{s}\) upward. Use the work-energy theorem to find (a) the rock's speed just as it left the ground and (b) its maximum height.

Short Answer

Expert verified
Initial speed: 29.2 m/s; Maximum height: 51.5 m.

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. This principle will guide us in solving the problem of finding both the initial speed and the maximum height of the rock.
02

Establish Energy Conservation Equation

When the rock is thrown, it has both kinetic energy and potential energy as it rises. The total mechanical energy (sum of kinetic and potential energy) will be conserved if we neglect air resistance. So, use the equation:\[K_i + U_i = K_f + U_f\]where \(K\) is kinetic energy \((K = \frac{1}{2}mv^2)\) and \(U\) is potential energy \((U = mgh)\). At ground level, initial potential energy \(U_i = 0\), and when it is 15 m above, use \(U_f = mgh\).
03

Determine the Mass of the Rock

Since the weight of the rock is 20 N and the acceleration due to gravity \(g = 9.81 \, m/s^2\), we can find the mass using the formula:\[m = \frac{W}{g} = \frac{20 \, N}{9.81 \, m/s^2}\]Calculate the mass.
04

Calculate Initial Speed Using Known Conditions

Apply the conservation of energy equation from the ground to the point 15 m high:\[\frac{1}{2}mv_i^2 = \frac{1}{2}mv_f^2 + mgh\]Substitute \(v_f = 25 \, m/s\), \(h = 15 \, m\), and solve for \(v_i\).
05

Solve for Maximum Height Using Energy Conservation

At maximum height, the final speed of the rock is zero \((v_f = 0)\). Use the known initial kinetic energy to find the potential energy at maximum height:\[\frac{1}{2}mv_i^2 = mgH_{max}\]Solve for \(H_{max}\) by substituting the calculated initial velocity \(v_i\) from Step 4.
06

Calculate and Verify Solutions

Perform the calculations for \(v_i\) and \(H_{max}\) using the given and derived data: the mass \(m\), gravitational acceleration \(g\), height \(h\), and observed speed \(v_f\). Ensure the values satisfy the energy conservation equation throughout the process.

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Key Concepts

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

Kinetic Energy
Kinetic energy refers to the energy that an object possesses due to its motion. It's expressed mathematically as \(K = \frac{1}{2}mv^2\), where \(m\) is the mass of the object and \(v\) is its velocity. This formula tells us that an increase in either mass or velocity will result in a higher kinetic energy.
When you throw a rock up into the air, it initially has high kinetic energy due to its initial speed. As it rises, this energy is gradually converted to potential energy. At the point when the rock is at 15 meters and moving at 25 m/s, we can calculate its kinetic energy using the formula.
Kinetic energy is vital in determining how far and how high the rock will go. Understanding how kinetic energy transforms is foundational to applying the work-energy theorem.
Potential Energy
Potential energy is the energy stored due to an object's position or height. For objects near the Earth's surface, it's calculated by the formula \(U = mgh\), where \(m\) is mass, \(g\) is gravitational acceleration (approximately \(9.81 \, m/s^2\)), and \(h\) is height above the reference point.
In the context of throwing the rock, potential energy increases as the rock ascends. At ground level, potential energy is minimal since height \(h = 0\). However, as the rock reaches 15 meters, it surely gains potential energy. This increase comes at the expense of its kinetic energy.
The potential energy at a given point helps us understand how high the rock could theoretically climb if no more kinetic energy remains. This concept is crucial in solving for the maximum height the rock reaches.
Energy Conservation
The principle of energy conservation is the cornerstone of physics related to any motion. It states that energy can neither be created nor destroyed, only transformed from one form to another. In the absence of non-conservative forces, like air resistance, mechanical energy (the sum of kinetic and potential energy) remains constant.
For the rock problem, understanding energy conservation allows us to set up equations which relate the initial and final states of the rock's energy. Initially, the rock has a lot of kinetic energy and very little potential energy. As it gains height, energy conversion occurs, mirroring conservation.
Utilizing this principle, you can calculate parameters like initial launch speed and maximum height by balancing energy equations between different points of the rock's motion. This makes energy conservation an incredibly powerful tool in solving real-world physics problems.
Mechanical Energy
Mechanical energy is the total energy possessed by an object, comprising both kinetic and potential energy. In the partial absence of external forces like air resistance, this energy remains constant. Mechanical energy at any one stage of the rock's journey can be expressed as \(E = K + U\), where \(K\) is kinetic energy and \(U\) is potential energy.
Understanding mechanical energy helps us analyze the energy state of an object at different positions. For example, when the rock is firmly on the ground, kinetic energy dominates. However, as it ascends to 15 meters or any higher altitude, potential energy comes to the forefront.
Mechanical energy conservation provides a straightforward method for calculating changing states such as how the rock's speed or height vary, making it a useful concept in predicting and verifying the outcome of physical actions.

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Most popular questions from this chapter

CALC Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount \(x,\) a force along the \(x\) -axis with \(x\) -component \(F_{x}=k x-b x^{2}+c x^{3}\) must be applied to the free end. Here \(k=100 \mathrm{N} / \mathrm{m}, b=700 \mathrm{N} / \mathrm{m}^{2},\) and \(c=12,000 \mathrm{N} / \mathrm{m}^{3} .\) Note that \(x>0\) when the spring is stretched and \(x<0\) when it is compressed. (a) How much work must be done to stretch this spring by 0.050 m from its unstretched length? (b) How much work must be done to compress this spring by 0.050 m from its unstretched length? (c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of \(F_{x}\) on \(x\) . (Many real springs behave qualitatively in the same way.)

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