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A \(25.0\) -kg child on a \(2.00\) -m-long swing is released from rest when the ropes of the swing make an angle of \(30.0^{\circ}\) with the vertical. (a) Neglecting friction, find the child's speed at the lowest position. (b) If the actual speed of the child at the lowest position is \(2.00 \mathrm{~m} / \mathrm{s}\), what is the mechanical energy lost due to friction?

Short Answer

Expert verified
The child's speed at the lowest position of the swing, ignoring friction, is \(4.59 m/s\) and the mechanical energy lost due to friction is \(15.66 J\).

Step by step solution

01

Calculate the potential energy at the top of the swing

The potential energy (PE) at the top of the swing can be calculated using the formula \(PE = mgh\), where m is the mass, g is the gravity constant (9.8 m/s²), and h is the height. The height here can be calculated using trigonometry, as it is the vertical component of the length of the swing, i.e., \(h = L - Lcos(\theta)\) where L is the length of the swing and \(\theta\) is the angle. Substituting the given values in, we get \(h = 2 - 2*cos(30) = 2 - 2*(0.866) = 0.268\ m\). So, the potential energy is \(PE = 25*9.8*0.268 = 65.66 J\)
02

Equate the potential energy at the top to the kinetic energy at the bottom

At the bottom of the swing, all the potential energy is converted into kinetic energy (KE), if we neglect friction. As, by the law of conservation of energy, \(PE = KE\). This allows us to calculate the velocity at the bottom of the swing using the formula for kinetic energy \(KE = 1/2 * m * v^2\), where v is the velocity. Solving for velocity, we get \(v = sqrt((2*PE)/m) = sqrt((2*65.66)/25) = 4.59 m/s\).
03

Calculate the mechanical energy lost due to friction

If the actual speed of the child at the bottom position is \(2.00 m/s\), we can calculate the actual kinetic energy using the formula \(KE = 1/2 * m * v^2 = 1/2 * 25 * (2^2) = 50 J\). The energy lost due to friction is the difference between the initial potential energy and the final kinetic energy, i.e., \(Energy_{lost} = PE - KE = 65.66 - 50 = 15.66 J\).

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Key Concepts

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

Potential Energy Calculation
When studying the conservation of energy in physics, one of the key concepts is the calculation of potential energy (PE). The potential energy is the energy stored in an object due to its position or height above the ground. It is given by the formula
\( PE = mgh \)
where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximately \( 9.8 m/s^2 \) on Earth), and \( h \) is the height above the reference point.

In the context of our textbook exercise, potential energy is calculated at the top of a swing when a child is at the furthest point from the ground. The height \( h \) can be found using trigonometry, considering the swing's length and the angle with the vertical.

To make this clearer to students, consider the swing as a part of a circle's circumference; at a certain angle, a portion of the swing's length contributes to the height above the ground. This vertical component is essential in calculating potential energy and can be determined trigonometrically.
Kinetic Energy Calculation
Another foundational stone in studying energy transformation is the kinetic energy (KE) calculation. Kinetic energy is the energy an object possesses due to its motion. It's quantified using the formula
\( KE = \frac{1}{2}mv^2 \)
where \( m \) is the mass of the object and \( v \) is the velocity of the object. In our swing example, when the swing reaches its lowest point, all the potential energy (assuming no energy loss to friction) has been converted into kinetic energy.

Understanding this exchange is pivotal for students as it illustrates the conservation of mechanical energy principle—a fundamental concept in physics. The child’s velocity at the bottom of the swing can be calculated by rearranging the kinetic energy formula to solve for velocity, considering the potential energy at the top is now converted to kinetic energy at the bottom. This implies that any object in free fall will increase in velocity until all its potential energy is converted into kinetic energy.
Mechanical Energy Loss
In an ideal system without non-conservative forces, the mechanical energy—being the sum of potential and kinetic energy—remains constant. However, when non-conservative forces like friction are present, energy is not conserved within the system and is instead transformed into heat or sound, resulting in a loss of mechanical energy.

The textbook problem gives students a practical understanding of this with the child on the swing. The difference in the calculated maximum velocity and the actual velocity at the lowest point indicates the presence of frictional forces at work, such as air resistance and friction in the swing's pivot. By comparing the theoretical kinetic energy to the actual, we measure the mechanical energy lost. It’s a clear and quantitative demonstration that in real-world scenarios, various forms of energy dispersal occur, affecting the system’s total mechanical energy.
Trigonometry in Physics
Trigonometry isn't just an abstract branch of mathematics; it finds practical application in physics, especially when analyzing forces and motion in two or three dimensions. In our swing example, trigonometry helps determine the component of the swing's length that contributes to the height above the ground, which is critical for calculating potential energy.

The child’s swing forms a triangle with the ropes and the vertical, allowing us to use sine, cosine, or tangent to find missing lengths or angles. This application of trigonometry is a fundamental skill for students to learn as it will recur in multiple areas of physics, from projectile motion to analyzing vectors. Breaking down these trigonometric calculations into more manageable steps can be an effective strategy for improving student comprehension.

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Most popular questions from this chapter

A 2 \(300-\mathrm{kg}\) pile driver is used to drive a steel beam into the ground. The pile driver falls \(7.50 \mathrm{~m}\) before coming into contact with the top of the beam, and it drives the beam \(18.0 \mathrm{~cm}\) farther into the ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.

A flea is able to jump about \(0.5 \mathrm{~m}\). It has been said that if a flea were as big as a human, it would be able to jump over a 100-story building! When an animal jumps, it converts work done in contracting muscles into gravita_ional potential energy (with some steps in between). The maximum force exerted by a muscle is proportional to its cross-sectional area, and the work done by the muscle is this force times the length of contraction. If we magnified a flea by a factor of 1000, the cross section of its muscle would increase by \(1000^{2}\) and the length of contraction would increase by \(1000 .\) How high would this "superflea" be able to jump? (Don't forget that the mass of the "superflea" increases as well.)

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A horizontal spring attached to a wall has a force constant of \(850 \mathrm{~N} / \mathrm{m} . \mathrm{A}\) block of mass \(1.00 \mathrm{~kg}\) is attached to the spring and oscillates freely on a horizontal, frictionless surface as in Active Figure \(5.20 .\) The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? (b) What are the two points of interest? (c) Find the energy stored in the spring when the mass is stretched \(6.00 \mathrm{~cm}\) from equilibrium and again when the mass passes through cquilibrium after being released from rest. (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. (e) What is the speed at the halfway point? Why isn't it half the speed at equilibrium?

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