/*! 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 79 A 5.00-kg block is moving at \(\... [FREE SOLUTION] | 91Ó°ÊÓ

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A 5.00-kg block is moving at \(\upsilon_0\) \(=\) 6.00 m/s along a frictionless, horizontal surface toward a spring with force constant \(k\) = 500 N/m that is attached to a wall (\(\textbf{Fig. P6.79}\)). The spring has negligible mass. (a) Find the maximum distance the spring will be compressed. (b) If the spring is to compress by no more than 0.150 m, what should be the maximum value of \(\upsilon_0\)?

Short Answer

Expert verified
(a) Maximum compression is 0.6 m. (b) Maximum initial velocity is 1.5 m/s.

Step by step solution

01

Understanding the Energy Conservation

The block possesses kinetic energy as it moves towards the spring, which is then transferred into potential energy stored in the spring when compressed. Since the surface is frictionless, we can use energy conservation principles: the initial kinetic energy of the block equals the potential energy of the spring at maximum compression.
02

Formula for Kinetic Energy

The kinetic energy of the block is given by \( KE = \frac{1}{2}mv_0^2 \), where \( m = 5.00 \) kg and \( v_0 = 6.00 \) m/s.
03

Calculate Initial Kinetic Energy

Substitute the given values into the kinetic energy formula: \( KE = \frac{1}{2} \times 5.00 \times 6.00^2 = 90.0 \) J.
04

Formula for Potential Energy in the Spring

The potential energy stored in a compressed spring is given by \( PE = \frac{1}{2}kx^2 \), where \( k = 500 \) N/m and \( x \) is the maximum compression.
05

Equate Kinetic and Potential Energies

Set the kinetic energy equal to the potential energy to find \( x \): \( 90.0 = \frac{1}{2} \times 500 \times x^2 \).
06

Solve for Maximum Compression

Solve the equation for \( x \): \( x^2 = \frac{180}{500} \). Calculate \( x \) by taking the square root: \( x = \sqrt{0.36} = 0.6 \) m.
07

Limit Maximum Compression to 0.150 m

If the spring compresses no more than 0.150 m, set \( \frac{1}{2}kx^2 \) equal to the maximal kinetic energy, solve for the new \( v_0 \).
08

Solve for New Maximum Velocity

Using \( PE = \frac{1}{2}kx^2 = \frac{1}{2} \times 500 \times 0.150^2 = 5.625 \) J, solve the kinetic energy equation: \( 5.625 = \frac{1}{2} \times 5.00 \times v_0^2 \). Calculate \( v_0 = \sqrt{2 \times 5.625 \div 5.00} = 1.5 \) m/s.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of motion. Any object that is moving has kinetic energy, which depends on the object's mass and velocity. It's calculated using the formula:
\[KE = \frac{1}{2}mv^2\]where \( m \) is the mass of the object, and \( v \) is its velocity.
This energy is directly proportional to the mass of the object and the square of its velocity. Therefore, small changes in speed can have a big impact on kinetic energy. Imagine a 5 kg block moving at 6 m/s on a frictionless surface, like in our original problem. Using the kinetic energy formula, its energy is evaluated as 90 Joules.
Understanding kinetic energy helps predict how much work is done when something is stopped or how much energy is required to reach a certain speed. Its conservation is key in systems like the frictionless block and spring, where energy transforms from kinetic to potential and vice versa.
Potential Energy
Potential energy is stored energy based on position or configuration. For springs, this energy is stored when the spring is either compressed or stretched. The formula for potential energy in a spring is:
\[PE = \frac{1}{2}kx^2\]where \( k \) represents the spring constant, and \( x \) is how much the spring is compressed or stretched.
In our scenario, as the block compresses the spring, it transfers kinetic energy into potential energy stored in the spring. The potentials are equal at maximum compression because the environment is frictionless, meaning no energy is lost. If the block's initial kinetic energy is known, like the 90 Joules calculated earlier, we can determine the maximum compression by equating kinetic and potential energies.
Understanding potential energy allows us to calculate how much work is needed to compress a spring or how much energy will be released if the spring returns to its resting state.
Spring Constant
The spring constant is a measure of a spring's stiffness, denoted as \( k \). It indicates how much force is needed to compress the spring by a certain distance. A higher spring constant means a stiffer spring, requiring more force to compress or stretch.
In our problem, the spring constant is 500 N/m. This value helps us understand how the spring will behave when it is compressed by the moving block. The amount of compression relates directly to how much potential energy is stored, as shown in the potential energy formula.
Having a clear grasp of the spring constant not only tells us how springs behave but also how they interact with other moving items like our kinetic block. This concept plays a crucial role in many real-life applications, from vehicle suspensions to measuring equipment. It's the essence of understanding and designing systems where spring action is needed.

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Most popular questions from this chapter

A mass \(m\) slides down a smooth inclined plane from an initial vertical height \(h\), making an angle \(\alpha\) with the horizontal. (a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height \(h\). (b) Use the work\(-\)energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if the mass had been dropped from height \(h\), independent of the angle \(\alpha\) of the incline. Explain how this speed can be independent of the slope angle. (c) Use the results of part (b) to find the speed of a rock that slides down an icy frictionless hill, starting from rest 15.0 m above the bottom.

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