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Ch 11: Impulse and Momentum
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 11, Problem 79a

In Problems 76,77,78,76, 77, 78, and 7979 you are given the equation(s) used to solve a problem. For each of these, you are to write a realistic problem for which this is the correct equation(s).
12(0.30 kg)(0 m/s)2+12(3.0 N/m)(Δx2)2=12(0.30 kg)(v1x)2+12(3.0 N/m)(0 m)2\(\frac{1}{2}\) (0.30 \(\text{ kg}\)) (0 \(\text{ m/s}\))^2 + \(\frac{1}{2}\) (3.0 \(\text{ N/m}\)) (\(\Delta\) x_2)^2 = \(\frac{1}{2}\) (0.30 \(\text{ kg}\)) (v_{1x})^2 + \(\frac{1}{2}\) (3.0 \(\text{ N/m}\)) (0 \(\text{ m}\))^2

Verified step by step guidance
1
Step 1: Recognize the given equation as an application of the conservation of mechanical energy. The equation equates the total initial energy (kinetic energy + potential energy) to the total final energy (kinetic energy + potential energy).
Step 2: Identify the components of the equation. The term \( \frac{1}{2}(0.30 \text{ kg})(0 \text{ m/s})^2 \) represents the initial kinetic energy of a 0.30 kg object moving at 0 m/s. The term \( \frac{1}{2}(3.0 \text{ N/m})(\Delta x_2)^2 \) represents the initial elastic potential energy stored in a spring with a spring constant of 3.0 N/m and a compression/stretch of \( \Delta x_2 \).
Step 3: Similarly, \( \frac{1}{2}(0.30 \text{ kg})(v_{1x})^2 \) represents the final kinetic energy of the object moving with velocity \( v_{1x} \), and \( \frac{1}{2}(3.0 \text{ N/m})(0 \text{ m})^2 \) represents the final elastic potential energy of the spring when it is at its equilibrium position (no compression/stretch).
Step 4: Write a realistic problem based on this equation. For example: 'A 0.30 kg block is initially at rest on a frictionless surface and is attached to a spring with a spring constant of 3.0 N/m. The spring is compressed by a distance \( \Delta x_2 \) and then released. Determine the velocity of the block \( v_{1x} \) when the spring returns to its equilibrium position.'
Step 5: To solve the problem, use the conservation of mechanical energy principle. Substitute the known values into the equation: \( \frac{1}{2}(0.30)(0)^2 + \frac{1}{2}(3.0)(\Delta x_2)^2 = \frac{1}{2}(0.30)(v_{1x})^2 + \frac{1}{2}(3.0)(0)^2 \). Simplify and solve for \( v_{1x} \) in terms of \( \Delta x_2 \).

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Key Concepts

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

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 mechanical systems, this means that the total mechanical energy (kinetic plus potential) remains constant if only conservative forces are acting. This concept is crucial for understanding how energy is transferred and transformed in physical systems, such as in the given equation involving kinetic and potential energy.
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Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion, calculated using the formula KE = 1/2 mv², where m is the mass and v is the velocity of the object. In the context of the provided equation, the kinetic energy terms represent the energy of the object at different states of motion, which is essential for analyzing how energy changes as the object moves.
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Potential Energy in Springs

Potential energy in springs is described by Hooke's Law, which states that the potential energy stored in a spring is given by PE = 1/2 k(Δx)², where k is the spring constant and Δx is the displacement from the equilibrium position. This concept is vital for understanding how the energy stored in a spring can be converted into kinetic energy as the spring is compressed or stretched, as illustrated in the equation provided.
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Related Practice
Textbook Question

In Problems 76,77,78,76, 77, 78, and 7979 you are given the equation(s) used to solve a problem. For each of these, you are to finish the solution of the problem, including a pictorial representation.

12(0.30 kg)(0 m/s)2+12(3.0 N/m)(Δx2)2=12(0.30 kg)(v1x)2+12(3.0 N/m)(0 m)2\(\frac{1}{2}\) (0.30 \(\text{ kg}\)) (0 \(\text{ m/s}\))^2 + \(\frac{1}{2}\) (3.0 \(\text{ N/m}\)) (\(\Delta\) x_2)^2 = \(\frac{1}{2}\) (0.30 \(\text{ kg}\)) (v_{1x})^2 + \(\frac{1}{2}\) (3.0 \(\text{ N/m}\)) (0 \(\text{ m}\))^2

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Textbook Question

A 45 g projectile explodes into three pieces: a 20 g piece with velocity 25 î m/s, a 15 g piece with velocity −10 î + 10ĵ m/s, and a 10 g piece with velocity −15 î − 20ĵ m/s. What was the projectile's velocity just before the explosion?

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Textbook Question

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Textbook Question

A white ball traveling at 2.0m/s hits an equal-mass red ball at rest. The white ball is deflected by 25° and slowed to 1.5m/s. What percentage of the initial mechanical energy is lost in the collision?

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