/*! 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 95 Two objects with masses 5.00 \(\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two objects with masses 5.00 \(\mathrm{kg}\) and 2.00 \(\mathrm{kg}\) hang 0.600 \(\mathrm{m}\) above the floor from the ends of a cord 6.00 \(\mathrm{m}\) long passing over a frictionless pulley. Both objects start from rest. Find the maximum height reached by the 2.00 -kg object.

Short Answer

Expert verified
The 2.00 kg object reaches a maximum height of 1.20 m.

Step by step solution

01

Understand the System

We have two objects connected by a cord over a pulley. The first object has a mass of 5.00 kg and the second object has a mass of 2.00 kg. The pulley is frictionless, meaning no energy is lost. Initially, both objects start from rest.
02

Identify the Forces

The 5.00 kg object will exert a greater gravitational force than the 2.00 kg object, causing the 5.00 kg object to move downward and the 2.00 kg object to move upward.
03

Set up the Energy Equation

Use the principle of conservation of energy. Initially, the only energy in the system is potential energy from both masses. As the system moves, gravitational potential energy changes to kinetic energy.
04

Potential Energy Change

The change in potential energy when the 5.00 kg object moves down (and the 2.00 kg object moves up) is given by:\[ \Delta U = m_1 g h_1 - m_2 g h_2 \]where \(m_1 = 5.00 \, \mathrm{kg}\), \(m_2 = 2.00 \, \mathrm{kg}\), and \(h_1 = h_2 = x\) since both move the same distance.
05

Kinetic Energy Conversion

All the potential energy lost by the 5.00 kg object converts to kinetic energy in the system:\[ \Delta U = \Delta K = \frac{1}{2} (m_1 + m_2) v^2 \]Solve for \(v^2\) to find the velocity at equilibrium before stopping due to the 2.00 kg block reaching max height.
06

Calculate Maximum Height

After solving, use the velocity to find the new height of the 2.00 kg object using the energy conversion equations to find potential energy at this peak height:\[ m_2 g h_{max} = \frac{1}{2} m_1 g h + \frac{1}{2} m_2 g h \]
07

Solve the Equations

Substitute and solve for maximum height \(h_{max}\). You will find the 2.00 kg object reaches a height of 1.20 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 91Ó°ÊÓ!

Key Concepts

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

Pulley Systems
Pulley systems are fundamental in mechanics, often used to lift or move loads with reduced effort. In a simple pulley system like the one described in our exercise, two masses are connected by a cord over a pulley. The pulley's role is to change the direction of the force applied, allowing objects on either side to move in opposite directions. This pulley is frictionless, meaning it does not lose energy due to friction, making calculations simpler and focused on the transformation of energy. Pulley systems can be classified based on the number of wheels or pulleys used, like simple pulleys with one wheel or compound ones with multiple. The effectiveness of a pulley system is often measured by its mechanical advantage, which helps in determining how much easier a task is using the system compared to direct lifting. In our system, as one object goes up, the other comes down - this seesaw-like motion is integral to understanding how potential and kinetic energy interchange.
Gravitational Potential Energy
Gravitational potential energy is a type of energy an object possesses because of its position relative to the Earth's surface. When an object is at a height, it possesses energy that has the potential to do work because of the force of gravity acting on it.The formula for gravitational potential energy is given by:\[ U = mgh \]where:
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) at the Earth's surface),
  • \( h \) is the height above the reference point.
In our exercise, the potential energy of each mass changes as the 5.00 kg object descends and the 2.00 kg object ascends. This change in height results in a transfer of energy, showcasing the conservation principle where potential energy decreases in one object and the increase in another manifests as kinetic energy in the system overall.
Kinetic Energy
Kinetic energy is the energy an object possesses because of its motion. Whenever an object moves, it has kinetic energy, which depends on both its mass and velocity. The standard formula for kinetic energy is:\[ K = \frac{1}{2}mv^2 \]where:
  • \( m \) is the mass of the moving object,
  • \( v \) is the velocity of the object.
As the 5.00 kg object starts descending, its potential energy converts into kinetic energy, increasing the system's overall kinetic energy. The speed gained by the object is essentially the result of this energy conversion. In pulley systems, kinetic energy is shared between the connected objects. When the object with a greater mass descends, its lost potential energy is matched by a gain in kinetic energy, which in turn, provides energy to the smaller mass, allowing it to reach maximum height after the velocity energy has been fully optimized. This showcases how energy shifts within a system, and why understanding kinetic energy is crucial for predicting motions.

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

If the coefficient of static friction between a table and a uniform massive rope is \(\mu_{s},\) what fraction of the rope can hang over the edge of the table without the rope sliding?

A small remote-controlled car with mass 1.60 \(\mathrm{kg}\) moves at a constant speed of \(v=12.0 \mathrm{m} / \mathrm{s}\) in a vertical circle inside a hollow metal cylinder that has a radius of 5.00 \(\mathrm{m}(\) Fig. \(\mathrm{P} 5.120) .\) What is the magnitude of the normal force exerted on the car by the walls of the cylinder at (a) point \(A\) (at the bottom of the vertical circle) and (b) point \(B(\) at the top of the vertical circle)?

A steel washer is suspended inside an empty shipping crate from a light string attached to the top of the crate. The crate slides down a long ramp that is inclined at an angle of \(37^{\circ}\) above the horizontal. The crate has mass 180 \(\mathrm{kg}\) . You are sitting inside the crate (with a flashlight); your mass is 55 \(\mathrm{kg}\) . As the crate is sliding down the ramp, you find the washer is at rest with respect to the crate when the string makes an angle of \(68^{\circ}\) with the top of the crate. What is the coefficient of kinetic friction between the ramp and the crate?

A 25.0 -kg box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is \(0.25,\) and the coefficient of static friction is \(0.35 .\) (a) As the angle \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0 \(\mathrm{m}\) along the loading ramp?

A curve with a 120 -m radius on a level road is banked at the correct angle for a speed of 20 \(\mathrm{m} / \mathrm{s}\) . If an automobile rounds this curve at \(30 \mathrm{m} / \mathrm{s},\) what is the minimum coefficient of static friction needed between tires and road to prevent skidding?

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.