/*! 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 61 A \(250 \mathrm{~g}\) block is d... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(250 \mathrm{~g}\) block is dropped onto a relaxed vertical spring that has a spring constant of \(k=\) \(2.5 \mathrm{~N} / \mathrm{cm}\) (Fig. \(7-46)\). The block becomes attached to the spring and compresses the spring \(12 \mathrm{~cm}\) before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?

Short Answer

Expert verified
(a) 0.294 J; (b) -1.8 J; (c) 1.96 m/s; (d) 24 cm.

Step by step solution

01

Convert Units

Before solving, let's convert all measurements to SI units. The block's mass is given as 250 g, which converts to 0.25 kg. The spring constant is 2.5 N/cm, which is equivalent to 250 N/m when converted.
02

Calculate Work by Gravitational Force

The work done by the gravitational force can be calculated using the formula \( W_g = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity (9.81 m/s²), and \( h \) is the displacement (0.12 m). Thus, \( W_g = 0.25 \times 9.81 \times 0.12 \approx 0.294 \) J.
03

Calculate Work by Spring Force

The work done by the spring force is calculated using the formula \( W_s = -\frac{1}{2}kx^2 \), where \( k \) is the spring constant (250 N/m) and \( x \) is the compression (0.12 m). Thus, \( W_s = -\frac{1}{2}\times 250 \times (0.12)^2 = -1.8 \) J. The negative sign indicates that the spring does work against the block.
04

Apply Energy Conservation for Impact Speed

Using the conservation of mechanical energy, the initial gravitational potential energy is equal to the sum of the kinetic energy just before the impact and the work done by the spring force. Hence, \( mgh = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \). Calculate \( v \) using this equation: \( 0.25 \times 9.81 \times 0.12 = \frac{1}{2} \times 0.25 \times v^2 + 1.8 \). Solving for \( v \), \( v \approx 1.96 \) m/s.
05

Calculate Maximum Compression with Doubled Speed

If the impact speed is doubled, \( v = 2 \times 1.96 = 3.92 \) m/s. Using energy conservation again: \( \frac{1}{2}mv_i^2 = mgh + \frac{1}{2}kx^2 \), plug in the values: \( \frac{1}{2} \times 0.25 \times (3.92)^2 = 0.25 \times 9.81 \times x + \frac{1}{2} \times 250 \times x^2 \). Solve this quadratic equation for the new compression \( x \). The solution to this equation provides \( x \approx 24 \) cm.

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Key Concepts

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

Gravitational Force
Gravitational force is an essential concept when studying work and energy, particularly in the context we have here. It is the force of attraction between a mass and the Earth, acting downward towards the center of the Earth. The work done by gravitational force, therefore, can be calculated using the formula: \[ W_g = m imes g imes h \]
  • \( m \): the mass of the object, measured in kilograms (kg).
  • \( g \): the acceleration due to gravity, approximately \(9.81 \,\text{m/s}^2\) on Earth's surface.
  • \( h \): the vertical displacement or the height from which the object falls, in meters (m).
In our exercise, we find that the block has a mass of \(0.25 \, \text{kg}\) and falls a height of \(0.12 \, \text{m}\). Substituting these values into our formula, the gravitational work is found to be \(0.294 \, \text{J}\). This calculation shows how gravitational energy is converted into mechanical work, providing a fundamental understanding necessary for analyzing work and energy in physics.
Spring Force
The spring force operates under Hooke's Law, which states that the force exerted by a spring is directly proportional to its compression or stretch. The equation for the force by the spring is given by: \[ F_s = -k imes x \]
  • \( k \): the spring constant, denoting the stiffness of the spring, typically in \(\text{N/m}\).
  • \( x \): the compression or elongation of the spring from its natural length, in meters.
The work done by the spring force can be calculated using the formula: \[ W_s = -\frac{1}{2} k x^2 \] Here, the negative sign indicates that the work done by the spring force is against the direction of displacement of the block. In the context of our exercise, the spring constant is \(250 \, \text{N/m}\), and the spring compresses by \(0.12 \, \text{m}\). This calculation results in work done by the spring force of \(-1.8 \, \text{J}\). The energy stored is potential energy within the spring, playing a key role in the energy conservation of mechanical systems.
Energy Conservation
Energy conservation is a fundamental principle stating that energy cannot be created or destroyed, only transformed from one form to another. In the case of the block and spring system, the energy transforms from gravitational potential energy to kinetic energy, and finally to spring potential energy. The principle of mechanical energy conservation is stated as: \[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \]
  • Initial Gravitational Energy (\(mgh\)): Energy gained from height.
  • Kinetic Energy (\(\frac{1}{2}mv^2\)): Energy due to motion just before impact.
  • Spring Potential Energy (\(\frac{1}{2}kx^2\)): Energy stored in the spring at maximum compression.
Using these energies, we found the speed of the block just before the spring compression using the energy conservation principle, calculated as \(1.96 \, \text{m/s}\). This approach demonstrates how energy transformation within a closed system maintains total energy.
Compression of Spring
The compression of a spring is an important measure of how much energy is stored within it. In our exercise, after accounting for an increased initial speed, the behavior of the block and spring system changes due to energy conservation. Initially, the block had a certain potential and kinetic energy, which, upon impact, interacted with the spring's resistance and caused compression. With the impact speed doubled to \(3.92 \, \text{m/s}\), we revisit our energy equations: \[ \frac{1}{2}mv_i^2 = mgh + \frac{1}{2}kx^2 \] Solving the quadratic equation formed gives a new maximum compression for the spring. This demonstrates how doubling the kinetic energy of the block alters the mechanics of the system, resulting in a larger compression of approximately \(0.24 \, \text{m}\). Understanding the relationship between speed, energy, and compression helps students grasp the dynamic interactions within mechanical systems and the nature of spring-based energy storage.

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Most popular questions from this chapter

A breadbox is made to move along an \(x\) axis from \(x=0.15 \mathrm{~m}\) to \(x=1.20 \mathrm{~m}\) by a force with a magnitude given by \(F=\exp \left(-2 x^{2}\right),\) with \(x\) in meters and \(F\) in newtons. (Here exp is the exponential function.) How much work is done on the breadbox by the force?

(a) At a certain instant, a particle-like object is acted on by a force \(\vec{F}=(4.0 \mathrm{~N}) \hat{\mathrm{i}}-(2.0 \mathrm{~N}) \hat{\mathrm{j}}+(9.0 \mathrm{~N}) \hat{\mathrm{k}}\) while the object's velocity is \(\vec{v}=-(2.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}} .\) What is the instantaneous rate at which the force does work on the object? (b) At some other time, the velocity consists of only a \(y\) component. If the force is unchanged and the instantaneous power is \(-12 \mathrm{~W},\) what is the velocity of the object?

An initially stationary \(2.0 \mathrm{~kg}\) object accelerates horizontally and uniformly to a speed of \(10 \mathrm{~m} / \mathrm{s}\) in \(3.0 \mathrm{~s}\). (a) In that \(3.0 \mathrm{~s}\) interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?

In 1975 the roof of Montreal's Velodrome, with a weight of \(360 \mathrm{kN},\) was lifted by \(10 \mathrm{~cm}\) so that it could be centered. How much work was done on the roof by the forces making the lift? (b) In 1960 a Tampa, Florida, mother reportedly raised one end of a car that had fallen onto her son when a jack failed. If her panic lift effectively raised \(4000 \mathrm{~N}\) (about \(\frac{1}{4}\) of the car's weight) by \(5.0 \mathrm{~cm},\) how much work did her force do on the car?

A force \(\vec{F}\) in the positive direction of an \(x\) axis acts on an object moving along the axis. If the magnitude of the force is \(F=10 e^{-x / 2.0} \mathrm{~N}\), with \(x\) in meters, find the work done by \(\vec{F}\) as the object moves from \(x=0\) to \(x=2.0 \mathrm{~m}\) by \((\) a \()\) plotting \(F(x)\) and estimating the area under the curve and (b) integrating to find the work analytically.

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