/*! 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 29 A vertical spring with a spring ... [FREE SOLUTION] | 91Ó°ÊÓ

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A vertical spring with a spring constant of \(450 \mathrm{N} / \mathrm{m}\) is mounted on the floor. From directly above the spring, which is unstrained, a \(0.30-\mathrm{kg}\) block is dropped from rest. It collides with and sticks to the spring, which is compressed by \(2.5 \mathrm{cm}\) in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in \(\mathrm{cm}\) ) above the compressed spring was the block dropped?

Short Answer

Expert verified
4.8 cm above the spring.

Step by step solution

01

Understanding the Problem

To solve this problem, we need to determine the height from which the block was dropped. The potential energy of the block at the height equals the elastic potential energy when the block compresses the spring.
02

Converting Units

The spring compression is given in centimeters, so we need to convert it to meters for consistency in our calculations with other SI units. Therefore, the compression is \[x = 2.5 \text{ cm} = 0.025 \text{ m}. \]
03

Identifying the Energy Transformation

Initially, the block has gravitational potential energy which is converted into the spring's elastic potential energy when the block compresses the spring. We use the formula for gravitational potential energy \[U_g = m g h,\]and for elastic potential energy \[U_e = \frac{1}{2} k x^2.\]
04

Setting Potential Energies Equal

At the moment the spring is fully compressed, the gravitational potential energy is completely converted into elastic potential energy. Therefore, set \[m g h = \frac{1}{2} k x^2,\]and solve for the height \( h \).
05

Solving for Height

Substitute the known values into the equation. We have:\[0.30 \cdot 9.8 \cdot h = \frac{1}{2} \cdot 450 \cdot (0.025)^2.\]Simplify and solve for \( h \):\[0.3 \cdot 9.8 \cdot h = \frac{1}{2} \cdot 450 \cdot 0.000625,\]\[2.94h = 0.140625,\]\[h = \frac{0.140625}{2.94} \approx 0.048 \text{ meters}.\]
06

Converting Height to Centimeters

Convert the height back to centimeters to answer the question in the required unit:\[h = 0.048 \text{ m} \times 100 \frac{\text{cm}}{\text{m}} = 4.8 \text{ cm}.\]

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Key Concepts

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

Gravitational Potential Energy
Gravitational Potential Energy is the energy stored in an object due to its position in a gravitational field. When you lift an object off the ground, you give it gravitational potential energy. The energy is determined by the object's mass, the height it is raised, and the strength of the gravitational field. The formula for gravitational potential energy is \[U_g = mgh\],where:
  • \(U_g\) is the gravitational potential energy.
  • \(m\) is the mass of the object in kilograms.
  • \(g\) is the acceleration due to gravity, approximately \(9.8 \, \text{m/s}^2\) on Earth.
  • \(h\) is the height in meters.
This energy converts to other forms when the object moves, such as when a block falls toward a spring. As it falls, gravitational potential energy decreases, transforming into other energy forms like kinetic energy or elastic potential energy once it hits and compresses a spring.
Understanding this concept helps in analyzing how energies transform in physical systems, like in this case, where the energy at height becomes the energy that compresses a spring.
Elastic Potential Energy
Elastic Potential Energy is the energy stored in elastic materials as a result of their stretching or compressing. When an object like a spring is compressed or stretched, it stores energy that can be released later. The energy depends on both the amount of stretch/compression and the spring’s stiffness, known as the spring constant.
The formula for calculating elastic potential energy is:\[U_e = \frac{1}{2} k x^2\]where:
  • \(U_e\) is the elastic potential energy.
  • \(k\) is the spring constant, which indicates how stiff the spring is.
  • \(x\) is the displacement from the equilibrium position.
In our exercise, when the block hits the spring and compresses it, the gravitational potential energy converts into elastic potential energy. This stored energy is what slows down the block, bringing it momentarily to rest. Understanding this concept allows one to see how systems store and release energy, such as springs in clocks or shock absorbers in vehicles.
Spring Constant
The Spring Constant is a measure of a spring's stiffness. It provides an indication of how much force you need to apply to compress or stretch the spring by a certain amount. Denoted by \(k\), it is expressed in units of Newtons per meter (\(\text{N/m}\)).
In Hooke's Law, the spring constant relates the force exerted on the spring and the displacement it causes:\[F = kx\]where:
  • \(F\) is the force applied to the spring.
  • \(k\) is the spring constant.
  • \(x\) is the displacement of the spring from its initial position.
A larger spring constant means a stiffer spring, requiring more force for the same amount of displacement. For instance, in this exercise, a spring constant of 450 N/m means the spring is relatively stiff, requiring a significant force for compression. Understanding the spring constant is crucial for calculating elastic potential energy and interpreting how physical systems behave, especially in cases involving energy transformations like the one in our block and spring scenario.

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Most popular questions from this chapter

In 0.750 s, a 7.00-kg block is pulled through a distance of 4.00 m on a frictionless horizontal surface, starting from rest. The block has a constant acceleration and is pulled by means of a horizontal spring that is attached to the block. The spring constant of the spring is 415 N/m. By how much does the spring stretch?

A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 59 N/m and 33 N/m. Find the angle that the rod makes with the horizontal.

To measure the static friction coefficient between a 1.6-kg block and a vertical wall, the setup shown in the drawing is used. A spring (spring constant = 510 N/m) is attached to the block. Someone pushes on the end of the spring in a direction perpendicular to the wall until the block does not slip downward. The spring is compressed by 0.039 m. What is the coefficient of static friction?

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is \(1.25 \mathrm{m} / \mathrm{s},\) and its maximum acceleration is \(6.89 \mathrm{m} / \mathrm{s}^{2} .\) How much time elapses between an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum?

Two physical pendulums (not simple pendulums) are made from meter sticks that are suspended from the ceiling at one end. The sticks are uniform and are identical in all respects, except that one is made of wood (mass \(=0.17 \mathrm{kg}\) ) and the other of metal (mass \(=0.85 \mathrm{kg}\) ). They are set into oscillation and execute simple harmonic motion. Determine the period of (a) the wood pendulum and (b) the metal pendulum.

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