/*! 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 40 A 4.00-kg block of ice is placed... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 40

A 4.00-kg block of ice is placed against a horizontal spring that has force constant \(k=200 \mathrm{N} / \mathrm{m}\) and is compressed 0.025 \(\mathrm{m}\) . The spring is released and accelerates the block along a horizontal surface. You can ignore friction and the mass of the spring. (a) Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length. (b) What is the speed of the block after it leaves the spring?

Short Answer

Expert verified
Work done is 0.0625 Joules; speed of the block is 0.177 m/s.

Step by step solution

01

Understand the Problem

We need to calculate the work done by the spring on a 4.00-kg block when the spring is released from a compressed state. We also need to find the speed of the block when the spring returns to its natural length.
02

Use Hooke's Law to Find Potential Energy

Hooke's Law describes the potential energy stored in a spring as \[ PE = \frac{1}{2} k x^2 \]where \(k\) is the spring constant and \(x\) is the compression distance. Use this formula to find the potential energy stored in the spring.
03

Substitute Values for Energy Calculation

Substitute the given values into the potential energy equation: \[ PE = \frac{1}{2} \times 200 \, \text{N/m} \times (0.025 \, \text{m})^2 \]Calculate this energy.
04

Calculate the Work Done by the Spring

The work done by the spring is equal to the potential energy stored in it because all of this energy is transferred to the block. Calculate the value of the work using the expression derived in Step 3.
05

Use Energy Conservation to Find Speed

Since there is no friction, energy conservation tells us that the kinetic energy of the block equals the potential energy initially stored in the spring:\[ KE = \frac{1}{2} m v^2 = PE \]where \(m\) is the mass of the block and \(v\) is its velocity. Rearrange this equation to solve for \(v\).
06

Solve for Velocity

Substitute the mass of the block and the calculated potential energy/work into the rearranged energy equation:\[ v = \sqrt{\frac{2 \times PE}{m}} \]Calculate this velocity.

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Key Concepts

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

Hooke's Law
When dealing with springs, Hooke's Law is essential to understand how they work. This law states that the force exerted by a spring is directly proportional to its extension or compression. The expression is given by \( F = -kx \), where:
  • \( F \) is the force exerted by the spring
  • \( k \) is the spring constant, a measure of the spring's stiffness
  • \( x \) is the displacement from the spring's equilibrium position

The negative sign indicates that the spring force is a restoring force, acting in the opposite direction of the displacement. The potential energy stored in a compressed or stretched spring, according to Hooke's Law, is expressed as \( PE = \frac{1}{2} k x^2 \). This potential energy is what powers the motion of the spring when released.
Potential Energy
Potential energy is a form of energy that is stored in an object due to its position or configuration. In the case of a spring, its potential energy depends on how much it is compressed or stretched from its natural length. The formula for the potential energy stored in a spring is \( PE = \frac{1}{2} k x^2 \).
  • \( k \) is the spring constant, which tells us how stiff the spring is.
  • \( x \) is the compression or elongation from the natural (rest) position.

This stored energy is entirely released when the spring returns to its original length. In the given problem, understanding potential energy helps us calculate how much energy is stored in the spring before it is released to move the block.
Energy Conservation
In physics, the principle of energy conservation states that energy cannot be created or destroyed, only transformed from one form to another. In a frictionless environment like this problem, the stored potential energy in the spring totally converts into the kinetic energy of the block as the spring is released.
  • The initial potential energy of the spring: \( PE_{initial} = \frac{1}{2} k x^2 \)
  • The kinetic energy of the block: \( KE = \frac{1}{2} mv^2 \)

By this conservation, the initial potential energy of the spring becomes the kinetic energy of the block allowing us to solve for the velocity of the block after it leaves the spring.
Kinetic Energy
Kinetic energy is the energy of motion that an object possesses. For the block in the problem, kinetic energy comes into play once the block is released from the spring. The expression for kinetic energy is \( KE = \frac{1}{2} mv^2 \), where:
  • \( m \) is the mass of the object
  • \( v \) is the object's velocity

After the spring releases the block, all the potential energy that was stored in the spring transforms into kinetic energy, propelling the block forward. By setting the kinetic energy equal to the spring's original stored potential energy, we can find the block's velocity: \( v = \sqrt{\frac{2 \times PE}{m}} \). This equation helps us determine how fast the block moves after leaving the influence of the spring.

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Most popular questions from this chapter

A 2.0 -kg box and a 3.0 -kg box on a perfectly smooth horizontal floor have a spring of force constant 250 \(\mathrm{N} / \mathrm{m}\) compressed between them. If the initial compression of the spring is \(6.0 \mathrm{cm},\) find the acceleration of each box the instant after they are released. Be sure to include free-body diagrams of each box as part of your solution.

A factory worker pushes a 30.0 -kg crate a distance of 4.5 \(\mathrm{m}\) along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25 . (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?

BIO Power of the Human Heart. The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 \(\mathrm{L}\) of blood. Assume that the work done by the heart is equal to the work required to lift this amount of blood a height equal to that of the average American woman \((1.63 \mathrm{m}) .\) The density (mass per unit volume) of blood is \(1.05 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) . (a) How much work does the heart do in a day? (b) What is the heart's power output in watts?

BIO All birds, independent of their size, must maintain a power output of \(10-25\) watts per kilogram of body mass in order to fly by flapping their wings. (a) The Andean giant hummingbird (Patagona gigas) has mass 70 \(\mathrm{g}\) and flaps its wings 10 times per second while hovering. Estimate the amount of work done by such a hummingbird in each wingbeat. (b) A 70 -kg athlete can maintain a power output of 1.4 \(\mathrm{kW}\) for no more than a few seconds; the steady power output of a typical athlete is only 500 \(\mathrm{W}\) or so. Is it possible for a human-powered aircraft to fly for extended periods by flapping its wings? Explain.

A typical flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Take the mass of the insect to be \(10 \mathrm{g},\) and assume the wings move an average downward distance of 1.0 \(\mathrm{cm}\) during each stroke. Assuming 100 downward strokes per second, estimate the average power output of the insect.
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