/*! 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 10 Block Dropped on a Spring A \(25... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 10

Block Dropped on a Spring A \(250 \mathrm{~g}\) block is dropped onto a relaxed vertical spring that has a spring constant of \(k=\) 2.5 N/cm (Fig. 10-32). 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) 6.03 m/s (d) 1.696 m

Step by step solution

01

Given Data and Unit Conversion

First, write down all given data and convert units if necessary.Mass of the block,\(m=250 \mathrm{~g}=0.25 \mathrm{~kg}\).Spring constant,\(k=2.5 \mathrm{N/cm}=250 \mathrm{~N/m}\).Compression of the spring,\(x=12 \mathrm{~cm}=0.12 \mathrm{~m}\).
02

Work Done by Gravitational Force (a)

Use the formula for work done by gravitational force: \( W_g = mgh \) where \( h \) is the height (same as compression of spring).\( W_g = 0.25 \mathrm{~kg} \times 9.8 \mathrm{~m/s^2} \times 0.12 \mathrm{~m} \)\( W_g = 0.294 \mathrm{~J} \)Thus, the work done by the gravitational force is \(0.294 \mathrm{~J}\).
03

Work Done by Spring Force (b)

Use the formula for work done by the spring force: \( W_s = - \frac{1}{2} k x^2 \).\( W_s = - \frac{1}{2} \times 250 \mathrm{~N/m} \times (0.12 \mathrm{~m})^2 \)\( W_s = -1.8 \mathrm{~J} \)Thus, the work done by the spring force is \( -1.8 \mathrm{~J} \).
04

Speed of Block Just Before Impact (c)

Apply conservation of energy principle: Initial potential energy + initial kinetic energy = Final potential energy + final kinetic energy (Which is zero at max compression).\( mgh + \frac{1}{2} m v^2 = \frac{1}{2} k x^2 \).Since initial potential energy by height is converted to spring potential energy:\( 0.25 \mathrm{~kg} \times 9.8 \mathrm{~m/s^2} \times 0.12 \mathrm{~m} = \frac{1}{2} \times 250 \mathrm{~N/m} \times (0.12 \mathrm{~m})^2 \)Rearrange and solve for \(v\):\( v = \sqrt{(k x^2 - 2 m g x)/m}\)\( v = \sqrt{(250 \times (0.12)^2 - 2 \times 0.25 \times 9.8 \times 0.12)/0.25}\)\( v \approx 6.03 \mathrm{~m/s}\)Thus, the speed of the block just before impact is approximately \(6.03 \mathrm{~m/s}\).
05

Maximum Spring Compression if Impact Speed Doubled (d)

If the speed at impact is doubled, use the conservation of energy to find the new compression:\( mgh + \frac{1}{2} m (2v)^2 = \frac{1}{2} k x_{max}^2 \).Replace with known values, then solve for \( x_{max} \):\( x_{max} = \sqrt{\left( 4mv^2+2mgh \right)/k} \).\( x_{max} = \sqrt{\large{ \left( 4 \times 0.25 \times 6.03^2 + 2 \times 0.25 \times 9.8 \times 0.12 \right)/250}} \).\( x_{max} \approx 1.696\mathrm{~m} \).Thus, the maximum compression is approximately \(1.696\mathrm{~m}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Work and Energy
Work and energy are fundamental concepts in physics, closely related to each other. When a force acts on an object causing it to move, work is done. The energy associated with this work can convert from one form to another. For instance, when a block drops and compresses a spring, its gravitational potential energy converts into kinetic energy and then into elastic potential energy stored in the spring. The work done by forces (like gravity and the spring force) helps determine the block's motion and compression of the spring.
Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed, only transformed. In the exercise, we apply this principle by equating the initial energy to the final energy.
When the block is dropped, its gravitational potential energy transforms into kinetic energy as it falls. Upon hitting the spring, this kinetic energy is then transferred into the spring as elastic potential energy. This means all the energies involved must add up to maintain a constant total energy.
  • Initial energy (gravitational potential energy + kinetic energy)
  • Final energy (elastic potential energy in the spring)
By setting these energies equal, we can solve for unknown variables, such as the speed of the block or the maximum compression of the spring.
Spring Constant
The spring constant, denoted as k, measures a spring's stiffness. A higher spring constant means a stiffer spring and larger force required to compress or extend it.
In the given exercise, the spring constant plays a crucial role in calculating the work done by the spring and the maximum compression. The spring's behavior is described by Hooke's Law:
  • The force exerted by the spring is proportional to its compression or extension (i.e., F = -kx)

Using this equation, we determine the work done by the spring, represented as
\( W_s = -\frac{1}{2} k x^2 \).
This work is necessary for figuring out how the energy is transformed and how much the spring compresses.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A Bear Slides A \(25 \mathrm{~kg}\) bear slides, from rest, \(12 \mathrm{~m}\) down a lodgepole pine tree, moving with a speed of \(5.6 \mathrm{~m} / \mathrm{s}\) just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?

A Woman Leaps Vertically A \(55 \mathrm{~kg}\) woman leaps vertically from a crouching position in which her center of mass is \(40 \mathrm{~cm}\) above the ground. As her feet leave the floor, her center of mass is \(90 \mathrm{~cm}\) above the ground; it rises to \(120 \mathrm{~cm}\) at the top of her leap. (a) As she is pressing down on the ground during the leap, what is the average magnitude of the force on her from the ground? (b) What maximum speed does she attain?

Block Sticks to Spring A \(700 \mathrm{~g}\) block is released from rest at height \(h_{1}\) above a vertical spring with spring constant \(k=400 \mathrm{~N} / \mathrm{m}\) and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring \(19.0 \mathrm{~cm}\). How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of \(h_{1} ?(\mathrm{~d})\) If the block were released from height \(2 h_{1}\) above the spring, what would be the maximum compression of the spring?

Two Snowy Peaks Two snowy peaks are \(850 \mathrm{~m}\) and \(750 \mathrm{~m}\) above the valley between them. A ski run extends down from the top of the higher peak and then back up to the top of the lower one, with a total length of \(3.2 \mathrm{~km}\) and an average slope of \(30^{\circ}\) (Fig. 10-49). (a) A skier starts from rest at the top of the higher peak. At what speed will he arrive at the top of the lower peak if he coasts without using ski poles? Ignore friction. (b) Approximately what coefficient of kinetic friction between snow and skis would make him stop just at the top of the lower peak?

Voyager 2 Spacecraft Voyager 2 (of mass \(m\) and speed \(v\) relative to the Sun) approaches the planet Jupiter (of mass \(M\) and speed \(V_{J}\) relative to the Sun) as shown in Fig. \(10-60 .\) The spacecraft rounds the planet and departs in the opposite direction. What is its speed, relative to the Sun, after this slingshot encounter, which can be analyzed as a collision? Assume \(v=12 \mathrm{~km} / \mathrm{s}\) and \(V_{J}=13 \mathrm{~km} / \mathrm{s}\) (the orbital speed of Jupiter). The mass of Jupiter is very much greater than the mass of the spacecraft \((M \gg m)\).
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Turning Points in Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Translational Dynamics

Read Explanation

Linear Momentum

Read Explanation

Scientific Method Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.