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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 79b

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

Verified step by step guidance
1
Identify the type of problem: This is a conservation of energy problem where the total mechanical energy (kinetic + potential) is conserved. The equation provided represents the initial and final energy states of a system involving a spring and a moving object.
Break down the terms in the equation: The left-hand side represents the initial energy, which includes the spring's potential energy (1/2 k Δx²) and the object's kinetic energy (1/2 m v²). The right-hand side represents the final energy, which includes the object's kinetic energy and the spring's potential energy.
Simplify the equation: Substitute the given values into the equation. For the initial state, the spring's compression or extension (Δx) is unknown, and the object's velocity is 0 m/s. For the final state, the spring is at its equilibrium position (Δx = 0), and the object's velocity is v₁ₓ.
Rearrange the equation to solve for the unknown: Since the spring's potential energy in the final state is zero, the equation simplifies to 1/2 k Δx² = 1/2 m v₁ₓ². Solve for Δx or v₁ₓ depending on the question's requirement.
Interpret the result: Once the unknown is solved, you can create a pictorial representation of the system. This could include a diagram of the spring and object at their initial and final states, showing the compression/extension of the spring and the object's velocity.

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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 solving problems involving motion and forces, as it allows us to equate the initial and final energy states of a system.
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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. Understanding kinetic energy is essential for analyzing how the speed of an object changes as it moves under the influence of forces, particularly in problems involving collisions or energy transformations.
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Potential Energy in Springs

The potential energy stored in a spring is given by the formula PE = 1/2 k(Δx)², where k is the spring constant and Δx is the displacement from the spring's equilibrium position. This concept is vital for problems involving springs, as it allows us to calculate the energy stored in the spring when it is compressed or stretched, which can then be related to the kinetic energy of an object when the spring is released.
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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 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

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Textbook Question

A 20 kg wood ball hangs from a 2.0-m-long wire. The maximum tension the wire can withstand without breaking is 400 N. A 1.0 kg projectile traveling horizontally hits and embeds itself in the wood ball. What is the greatest speed this projectile can have without causing the wire to break?

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Textbook Question

A 1000 kg cart is rolling to the right at 5.0 m/s. A 70 kg man is standing on the right end of the cart. What is the speed of the cart if the man suddenly starts running to the left with a speed of 10 m/s relative to the cart?

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Textbook Question

A 2100 kg truck is traveling east through an intersection at 2.0 m/s when it is hit simultaneously from the side and the rear. (Some people have all the luck!) One car is a 1200 kg compact traveling north at 5.0 m/s. The other is a 1500 kg midsize traveling east at 10 m/s. The three vehicles become entangled and slide as one body. What are their speed and direction just after the collision?

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Textbook Question

A spaceship of mass 2.0×10⁶ kg is cruising at a speed of 5.0×10⁶ m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 5.0×10⁵ kg, is blown straight backward with a speed of 2.0×10⁶ m/s . A second piece, with mass 8.0×10⁵ kg, continues forward at 1.0×10⁶ m/s. What are the direction and speed of the third piece?

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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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