Ch. 08 - Conservation of Energy

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 08 - Conservation of Energy

Problem 23a

Chapter 8, Problem 23a

The 9.0-kg mass in Fig. 8–36 is held just barely in contact with a spring for which k = 450 N/m . When that mass is released, it falls, compressing the spring and pulling the 3.0-kg mass up. How far does the 9.0-kg mass fall before momentarily coming to rest? Ignore friction in the pulley.

Verified step by step guidance

Identify the key concepts involved: This problem involves energy conservation (gravitational potential energy, elastic potential energy) and the relationship between the two masses connected by the pulley system. The spring constant (k) and the masses are given, and we are tasked with finding the distance the 9.0-kg mass falls before coming to rest.

Set up the energy conservation equation: The total mechanical energy of the system is conserved. Initially, the 9.0-kg mass has gravitational potential energy, and the spring has no elastic potential energy. At the point where the 9.0-kg mass momentarily comes to rest, all the gravitational potential energy lost by the 9.0-kg mass is converted into elastic potential energy of the spring. The equation is: m1gh = 12kx2

Relate the displacement of the two masses: Since the two masses are connected by a pulley, the distance the 9.0-kg mass falls (h) is equal to the distance the spring is compressed (x). Thus, h = x. Substitute this into the energy conservation equation.

Solve for the displacement (x): Substitute the known values into the equation. The mass of the 9.0-kg object is m₁ = 9.0 kg, the spring constant is k = 450 N/m, and g = 9.8 m/s². The equation becomes: 9.0 × 9.8 × x = 12 × 450 × x2

Simplify and solve the quadratic equation: Rearrange the equation into standard quadratic form: 225x2 - 88.2x = 0 . Factorize or use the quadratic formula to solve for x. Discard any negative solution, as distance cannot be negative.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hooke's Law

Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position, expressed as F = -kx, where F is the force, k is the spring constant, and x is the displacement. In this scenario, the spring constant (k = 450 N/m) will determine how much the spring compresses when the 9.0-kg mass falls.

Gravitational Potential Energy

Gravitational potential energy (U) is the energy an object possesses due to its position in a gravitational field, calculated as U = mgh, where m is mass, g is the acceleration due to gravity (approximately 9.81 m/s²), and h is the height. As the 9.0-kg mass falls, its potential energy is converted into kinetic energy and then into spring potential energy when the spring is compressed.

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In this problem, the gravitational potential energy of the falling mass is converted into the elastic potential energy of the compressed spring, allowing us to set up an equation to find the distance the mass falls before momentarily coming to rest.

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