/*! 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 46 A 5.00 -kg ball of clay is throw... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 5.00 -kg ball of clay is thrown downward from a height of \(3.00 \mathrm{~m}\) with a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) onto a spring with \(k=\) \(1600 . \mathrm{N} / \mathrm{m} .\) The clay compresses the spring a certain maximum amount before momentarily stopping a) Find the maximum compression of the spring. b) Find the total work done on the clay during the spring's compression.

Short Answer

Expert verified
Answer: To find the maximum compression of the spring (x), use the formula: $$ x = \sqrt{\frac{2(PE_i + KE_i)}{k}} $$ where \(PE_i\) is the initial potential energy of the clay, calculated as \(mgh\), and \(KE_i\) is the initial kinetic energy, calculated as \(\frac{1}{2}mv_i^2\). Then, to find the total work done on the clay during the spring's compression (W_total), use the formula: $$ W_{total} = PE_s - (PE_i + KE_i) $$ where \(PE_s\) is the elastic potential energy stored in the spring, calculated as \(\frac{1}{2}kx^2\). The negative sign indicates that the work is done against the motion of the clay to stop it.

Step by step solution

01

Calculate the initial potential energy and kinetic energy of the clay

Before the clay falls, it has potential energy due to its height and kinetic energy due to its initial speed. We can find both of these using the following formulas: $$ PE_i = mgh $$ $$ KE_i = \frac{1}{2}mv_i^2 $$ where \(PE_i\) is the initial potential energy, \(m\) is the mass of the clay, \(g\) is the acceleration due to gravity (approximately \(9.81ms^{-2}\)), \(h\) is the initial height, \(KE_i\) is the initial kinetic energy, and \(v_i\) is the initial speed of the clay.
02

Write the energy balance equation

When the clay compresses the spring to its maximum compression, it has transferred all its initial potential and kinetic energy into the elastic potential energy stored in the spring. So, we write the energy balance equation as follows: $$ PE_i + KE_i = PE_s $$ where \(PE_s\) is the elastic potential energy stored in the spring.
03

Calculate the maximum compression of the spring

We can use Hooke's law to calculate the elastic potential energy stored in the spring: $$ PE_s = \frac{1}{2}kx^2 $$ where k is the spring constant and x is the maximum compression of the spring. Substituting this into the energy balance equation from Step 2, we get: $$ PE_i + KE_i = \frac{1}{2}kx^2 $$ Solving for x, we get: $$ x = \sqrt{\frac{2(PE_i + KE_i)}{k}} $$ Plug in the values for \(PE_i\), \(KE_i\), and k from Steps 1 and 3, and calculate x.
04

Calculate the total work done on the clay during spring's compression

Because the conservation of mechanical energy holds during the process, the total work done on the clay during the spring's compression can be expressed as the change in mechanical energy: $$ W_{total} = PE_s - (PE_i + KE_i) $$ The negative sign indicates that the work is done against the motion of the clay to stop it. Calculate \(W_{total}\) using the values from Steps 1 and 3, and the formula for \(PE_s\) from Step 3.

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 91Ó°ÊÓ!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of motion. When an object is moving, it possesses kinetic energy, which depends on its mass and speed. For the clay in our exercise, its kinetic energy is determined at the moment it is thrown downward. We use the formula:
  • \( KE_i = \frac{1}{2}mv_i^2 \)
Here, \( m \) is the mass of the clay (5.00 kg), and \( v_i \) is its initial speed (5.00 m/s).
Substitute these values into the formula to calculate the initial kinetic energy.
The concept of kinetic energy is crucial because it tells us how energy is transitioning from motion into another form—in this case, the clay uses kinetic energy to compress the spring.
Spring Compression
Spring compression occurs when a spring is pushed together, storing potential energy. This energy is known as elastic potential energy, and it can be calculated using Hooke's Law. We express this stored energy using:
  • \( PE_s = \frac{1}{2}kx^2 \)
Here, \( k \) is the spring constant (1600 N/m), which measures the spring's stiffness, and \( x \) is the amount of compression.
When the clay hits the spring, the initial potential and kinetic energy transform into the elastic potential energy of the spring as it compresses.
Understanding spring compression is pivotal as it showcases how mechanical energy transitions into stored energy and accounts for energy conservation principles.
Energy Conservation
Energy conservation is a fundamental principle in physics that states that the total energy in a closed system remains constant. In this problem, all initial energies are converted without loss.
Initially, the clay has potential energy from its height and kinetic energy from its motion.
At the moment the spring is fully compressed, these energies are transformed into elastic potential energy.The energy conservation equation for this process is:
  • \( PE_i + KE_i = PE_s \)
This means the sum of initial potential and kinetic energy is equal to the spring's stored energy.
Energy conservation helps us understand how different types of energy interact and balance within a system, demonstrating that energy doesn't vanish but rather morphs into different forms.

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 uniform chain of total mass \(m\) is laid out straight on a frictionless table and held stationary so that one-third of its length, \(L=1.00 \mathrm{~m},\) is hanging vertically over the edge of the table. The chain is then released. Determine the speed of the chain at the instant when only one-third of its length remains on the table.

A 1.00 -kg block compresses a spring for which \(k=\) 100. \(\mathrm{N} / \mathrm{m}\) by \(20.0 \mathrm{~cm}\) and is then released to move across a horizontal, frictionless table, where it hits and compresses another spring, for which \(k=50.0 \mathrm{~N} / \mathrm{m}\). Determine a) the total mechanical energy of the system, b) the speed of the mass while moving freely between springs, and c) the maximum compression of the second spring.

A 0.500 -kg mass is attached to a horizontal spring with \(k=100 . \mathrm{N} / \mathrm{m}\). The mass slides across a frictionless surface. The spring is stretched \(25.0 \mathrm{~cm}\) from equilibrium, and then the mass is released from rest. a) Find the mechanical energy of the system. b) Find the speed of the mass when it has moved \(5.00 \mathrm{~cm}\). c) Find the maximum speed of the mass.

A high jumper approaches the bar at \(9.0 \mathrm{~m} / \mathrm{s}\). What is the highest altitude the jumper can reach, if he does not use any additional push off the ground and is moving at \(7.0 \mathrm{~m} / \mathrm{s}\) as he goes over the bar?

You use your hand to stretch a spring to a displacement \(x\) from its equilibrium position and then slowly bring it back to that position. Which is true? a) The spring's \(\Delta U\) is positive. b) The spring's \(\Delta U\) is negative. c) The hand's \(\Delta U\) is positive. d) The hand's \(\Delta U\) is negative. e) None of the above statements is true.

See all solutions

Recommended explanations on Physics Textbooks

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.