/*! 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 85 A 5.00 -kg block is moving at \(... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 85

A 5.00 -kg block is moving at \(v_{0}=6.00 \mathrm{m} / \mathrm{s}\) along a frictionless, horizontal surface toward a spring with force constant \(k=500 \mathrm{N} / \mathrm{m}\) that is attached to a wall (Fig. P6.85). The spring has negligible mass. \begin{equation} \begin{array}{l}{\text { (a) Find the maximum distance the spring will be compressed. }} \\ {\text { (b) If the spring is to compress by no more than } 0.150 \mathrm{m}, \text { what }} \\\ {\text { should be the maximum value of } v_{0} ?}\end{array} \end{equation}

Short Answer

Expert verified
(a) Maximum compression is 0.6 m. (b) Maximum initial velocity for 0.150 m compression is 2.74 m/s.

Step by step solution

01

Determine the Initial Energy of the System

The block's initial kinetic energy can be calculated using the formula for kinetic energy: \[ KE_i = \frac{1}{2}mv_0^2 \] where \( m = 5.00 \text{ kg} \) and \( v_0 = 6.00 \text{ m/s} \). Substitute these values into the equation to find \( KE_i \).
02

Set Up the Energy Conservation Equation

Since there is no friction, energy is conserved. The initial kinetic energy of the block will be converted completely into the potential energy stored in the spring at maximum compression. The potential energy in the spring is given by: \[ PE_s = \frac{1}{2}kx^2 \] Equate the kinetic energy to the potential energy to find: \[ \frac{1}{2}mv_0^2 = \frac{1}{2}kx^2 \] Simplify to solve for \( x \), the compression of the spring.
03

Solve for Maximum Compression

By solving the equation \( mv_0^2 = kx^2 \), we find:\[ x = \sqrt{\frac{mv_0^2}{k}} \] Substitute \( m = 5.00 \text{ kg} \), \( v_0 = 6.00 \text{ m/s} \), and \( k = 500 \text{ N/m} \) to solve for \( x \).
04

Calculate Potential Energy for Given Compression

If the compression \( x \) is limited to \( 0.150 \text{ m} \), calculate the total stored energy using:\[ PE_s = \frac{1}{2}k(0.150)^2 \] This energy will equal the maximum possible kinetic energy of the block.
05

Determine Maximum Velocity for Given Compression Limit

Using the maximum potential energy found in Step 4, solve for the maximum initial velocity \( v_0 \) using:\[ KE_i = \frac{1}{2}mv_{0_{max}}^2 = PE_s \] Solve for \( v_{0_{max}} \), keeping the compression limit in mind. Rearrange to find:\[ v_{0_{max}} = \sqrt{\frac{2PE_s}{m}} \] Substitute values to find \( v_{0_{max}} \).

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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 all about motion, and it describes how much energy an object has due to its movement. Imagine a ball rolling down a hill or a car speeding along a highway. That's kinetic energy at play! The formula to calculate kinetic energy is:
  • \( KE = \frac{1}{2}mv^2 \)
where \( m \) is the mass of the object, and \( v \) is its velocity.

In our exercise, we're looking at a 5.00-kg block moving with a velocity of 6.00 m/s. Substituting these values into the formula, we can find the initial kinetic energy of the block:

\[ KE_i = \frac{1}{2} \times 5.00 \times (6.00)^2 \] Calculating this gives us the initial kinetic energy, which serves as a starting point in our energy conservation analysis.
Potential Energy
Potential energy, on the other hand, is all about position. It measures the energy stored in an object because of its position or configuration. One of the common types of potential energy is the elastic potential energy found in springs.

For springs, the potential energy is given by:
  • \( PE_s = \frac{1}{2}kx^2 \)
where \( k \) is the spring constant (a measure of the spring's stiffness), and \( x \) is the compression or extension of the spring.

In this exercise, as the 5.00-kg block moves towards the spring and compresses it, its kinetic energy is transformed into elastic potential energy. By setting the initial kinetic energy equal to the maximum potential energy, we can determine the maximum distance the spring will compress.
Spring Compression
Spring compression is the action of a spring being pressed into a more compact form, storing potential energy in its coils. When a moving object, like our 5.00-kg block, hits the spring, the spring shortens and stores energy, which can be calculated using the potential energy formula for springs.

To find the maximum compression of the spring, we apply energy conservation principles. Initially, all energy is kinetic. As the block compresses the spring, this kinetic energy turns into potential energy. This transformation is captured by the equation:
  • \( \frac{1}{2}mv_0^2 = \frac{1}{2}kx^2 \)
Solving for \( x \), the maximum compression, involves substituting known values of mass, velocity, and spring constant into this equation:

\[ x = \sqrt{\frac{mv_0^2}{k}} \]Finally, with this formula, we can calculate how much the spring compresses when the block comes to a stop, effectively converting all its kinetic energy into stored potential energy.

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Most popular questions from this chapter

An ingenious bricklayer builds a device for shooting bricks up to the top of the wall where he is working. He places a brick on a vertical compressed spring with force constant \(k=450 \mathrm{N} / \mathrm{m}\) and negligible mass. When the spring is released, the brick is propelled upward. If the brick has mass 1.80 \(\mathrm{kg}\) and is to reach a maximum height of 3.6 \(\mathrm{m}\) above its initial position on the compressed spring, what distance must the bricklayer compress the spring initially? (The brick loses contact with the spring when the spring returns to its uncompressed length. Why?

BIO Chin-Ups, While doing a chin-up, a man lifts his body 0.40 \(\mathrm{m} .\) (a) How much work must the man do per kilogram of body mass? (b) The muscles involved in doing a chin-up can generate about 70 \(\mathrm{J}\) of work per kilogram of muscle mass. If the man can just barely do a \(0.40-\mathrm{m}\) chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical \(70-\mathrm{kg}\) man with 14\(\%\) body fat is about 43\(\%\) . (c) Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 \(\mathrm{J}\) of work per kilogram of muscle mass. (d) Adults and children have about the same percentage of muscle in their bodies. Explain why children can commonly do chin-ups more easily than their fathers.

CP A 20.0 -kg crate sits at rest at the bottom of a 15.0 -m-long ramp that is inclined at \(34.0^{\circ}\) above the horizontal. A constant horizontal force of 290 \(\mathrm{N}\) is applied to the crate to push it up the ramp. While the crate is moving, the ramp exerts a constant frictional force on it that has magnitude 65.0 \(\mathrm{N}\) . (a) What is the total work done on the crate during its motion from the bottom to the top of the ramp? (b) How much time does it take the crate to travel to the top of the ramp?

Half of a Spring. (a) Suppose you cut a massless ideal spring in half. If the full spring had a force constant \(k\) , what is the force constant of each half, in terms of \(k ?\) (Hint: Think of the original spring as two equal halves, each producing the same force as the entire spring. Do you see why the forces must be equal? (b) If you cut the spring into three equal segments instead, what is the force constant of each one, in terms of \(k ?\)

You throw a 20 -N rock vertically into the air from ground level. You observe that when it is 15.0 \(\mathrm{m}\) above the ground, it is trav- eling at 25.0 \(\mathrm{m} / \mathrm{s}\) upward. Use the work-energy theorem to find (a) the rock's speed just as it left the ground and (b) its maximum height.
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