/*! 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 21 A \(750.0 \mathrm{~kg}\) boulder... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(750.0 \mathrm{~kg}\) boulder is raised from a quarry \(125 \mathrm{~m}\) deep by a long chain having a mass of \(575 \mathrm{~kg} .\) This chain is of uniform strength, but at any point it can support a maximum tension no greater than 2.50 times its weight without breaking. (a) What is the maximum acceleration the boulder can have and still get out of the quarry, and (b) how long does it take for the boulder to be lifted out at maximum acceleration if it started from rest?

Short Answer

Expert verified
(a) The maximum acceleration of the boulder is 0.833 m/s². (b) It takes approximately 17.34 seconds to lift the boulder out at maximum acceleration.

Step by step solution

01

Calculate the weight of the boulder and the chain

The weight of the boulder is given by the equation \( W_b = m_b \cdot g \), where \( m_b \) is the mass of the boulder, and \( g \) is the acceleration due to gravity \( (9.81 \text{ m/s}^2) \). Thus, the weight of the boulder \( W_b = 750.0 \text{ kg} \times 9.81 \text{ m/s}^2 = 7357.5 \text{ N} \). Similarly, the weight of the chain \( W_c = 575.0 \text{ kg} \times 9.81 \text{ m/s}^2 = 5640.75 \text{ N} \).
02

Determine the maximum tension the chain can support

The chain can support a tension that is 2.50 times its weight without breaking. Thus, \( T_{max} = 2.50 \times W_c = 2.50 \times 5640.75 \text{ N} = 14101.875 \text{ N} \).
03

Set up the equation for maximum tension

The total force acting on the system, considering the maximum acceleration \( a \), is given by \( T_{max} = W_b + W_c + (m_b + m_c) \cdot a \). Rearranging gives the equation for \( a: a = \frac{T_{max} - (W_b + W_c)}{m_b + m_c} \). Where \( m_c = 575 \text{ kg} \).
04

Calculate the maximum acceleration

Substitute the known values into the equation: \( a = \frac{14101.875 \text{ N} - (7357.5 \text{ N} + 5640.75 \text{ N})}{750.0 \text{ kg} + 575.0 \text{ kg}} = \frac{14101.875 \text{ N} - 12998.25 \text{ N}}{1325.0 \text{ kg}} = 0.833 \text{ m/s}^2 \).
05

Calculate the time required to lift the boulder

Using the kinematic equation for constant acceleration \( d = \frac{1}{2} a t^2 \) and initial velocity zero, substitute \( d = 125 \text{ m} \) and \( a = 0.833 \text{ m/s}^2 \): \(125 = \frac{1}{2} \times 0.833 \times t^2 \), \( t^2 = \frac{125 \times 2}{0.833} \), \( t = \sqrt{\frac{250}{0.833}} \approx 17.34 \text{ s} \).

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.

Classical Mechanics
In classical mechanics, we explore the forces that govern the motion of objects in our everyday world. This branch of physics, developed by scientists like Isaac Newton, helps us understand how objects move and interact using laws and equations. One of the key contributions of classical mechanics is Newton’s Laws of Motion, which explain how forces affect motion and equilibrium.

### Newton’s Laws of Motion
- **First Law (Law of Inertia):** An object in motion stays in motion unless acted upon by a net external force. Similarly, an object at rest stays at rest.
- **Second Law (Law of Acceleration):** The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass, given by the equation: \( F = m imes a \).
- **Third Law (Action and Reaction):** For every action force, there is an equal and opposite reaction force.

These principles together provide a framework to solve problems involving motion, such as the lifting of a boulder out of a quarry. Understanding these laws allows us to calculate the required forces and motion parameters like tension and acceleration.
Kinematics
Kinematics is the study of motion without considering the forces that cause it. It focuses on the different aspects of motion such as speed, velocity, and acceleration. In the problem above, kinematics equations help us determine how an object moves under certain conditions.

### Key Kinematic Equations
- **Displacement (d):** The change in position of an object. It can be calculated for constant acceleration using the equation \( d = v_i imes t + \frac{1}{2} a imes t^2 \), where \( v_i \) is the initial velocity, \( t \) is time, and \( a \) is acceleration.
- **Velocity and Acceleration:** Velocity is the speed of an object in a given direction. Acceleration is the change of velocity over time.

For the boulder in our problem, we started with rest, which implies initial velocity, \( v_i \), is zero. We then applied the kinematic equation to find the time it takes for the boulder to be lifted with a constant maximum acceleration.
Forces and Tension
Forces are any interactions that cause an object to change its motion. Tension is a specific type of force that is transmitted through a string, cable, or chain when it is pulled tight by forces from opposite ends. In our scenario, both the weight of the chain and the boulder contribute to the total force that must be overcome for motion to occur.

### Understanding Tension
- **Weight as a Force:** The weight of an object is a force calculated by \( W = m imes g \), where \( g \) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\) on Earth).
- **Maximum Tension:** The chain can only hold a force up to a certain limit before it breaks. Here, it can withstand a tension up to 2.50 times its own weight.

In practice, solving for force and tension requires using their interrelatedness. The weight of the boulder and chain, coupled with the tension limit of the chain, provides the necessary conditions to calculate maximum acceleration.
Acceleration and Motion
Acceleration is the rate at which an object's velocity changes. Motion involves the activities or paths objects take when subject to forces and acceleration. In physics, understanding acceleration is crucial for predicting how and when an object will move.

### Calculating Acceleration
- **Derived Formula:** To find the maximum acceleration that doesn't exceed the tension capability of the chain, we rearrange the standard force equation to \( a = \frac{T_{max} - (W_b + W_c)}{m_b + m_c} \). Here, \( T_{max} \) is the maximum tension the chain can endure.
- **Constant Acceleration:** With a known force, objects reveal predictable paths and times required for travel. This aligns with the kinematic equation for objects at rest: \( d = \frac{1}{2} a t^2 \).

In this exercise, calculating the maximum acceleration ensures that the boulder can be safely lifted out of the quarry without risking the chain's integrity. This real-world application of acceleration examines practical thresholds to ensure efficient and safe motion.

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 block with mass \(m_{1}\) is placed on an inclined plane with slope angle \(\alpha\) and is connected to a second hanging block with mass \(m_{2}\) by a cord passing over a small, frictionless pulley (Figure 5.73). The coefficient of static friction is \(\mu_{\mathrm{s}}\) and the coefficient of kinetic fric- tion is \(\mu_{\mathrm{k}}\) (a) Find the mass \(m_{2}\) for which block \(m_{1}\) moves up the plane at constant speed once it is set in motion. (b) Find the mass \(m_{2}\) for which block \(m_{1}\) moves down the plane at constant speed once it is set in motion. (c) For what range of values of \(m_{2}\) will the blocks remain at rest if they are released from rest?

Shoes for the sports of bouldering and rock climbing are designed to provide a great deal of friction between the foot and the surface of the ground. On smooth rock these shoes might have a coefficient of static friction of 1.2 and a coefficient of kinetic friction of 0.90 . For a person wearing these shoes, what's the maximum angle (with respect to the horizontal) of a smooth rock that can be walked on without slipping? A. \(42^{\circ}\) B. \(50^{\circ}\) C. \(64^{\circ}\) D. Greater than \(90^{\circ}\)

A pickup truck is carrying a toolbox, but the rear gate of the truck is missing, so the box will slide out if it is set moving. The coefficients of kinetic and static friction between the box and the bed of the truck are 0.355 and \(0.650,\) respectively. Starting from rest, what is the shortest time in which this truck could accelerate uniformly to \(30.0 \mathrm{~m} / \mathrm{s}(\approx 60 \mathrm{mi} / \mathrm{h}) \quad\) without causing the box to slide? Include a free-body diagram of the toolbox as part of your solution. (Hint: First use Newton's second law to find the maximum acceleration that static friction can give the box, and then solve for the time required to reach \(30.0 \mathrm{~m} / \mathrm{s}\).)

An average person can reach a maximum height of about \(60 \mathrm{~cm}\) when jumping straight up from a crouched position. During the jump itself, the person's body from the knees up typically rises a distance of around \(50 \mathrm{~cm}\). To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. (a) With what initial speed does the person leave the ground to reach a height of \(60 \mathrm{~cm} ?\) (b) Make a free-body diagram of the person during the jump. (c) In terms of this jumper's weight \(\underline{W}\). what force does the ground exert on him or her during the jump?

A person pushes on a stationary \(125 \mathrm{~N}\) box with \(75 \mathrm{~N}\) at \(30^{\circ}\) below the horizontal, as shown in Figure 5.61 . The coefficient of static friction between the box and the horizontal floor is 0.80 . (a) Make a free-body diagram of the box. (b) What is the normal force on the box? (c) What is the friction force on the box? (d) What is the largest the friction force could be? (e) The person now replaces his push with a \(75 \mathrm{~N}\) pull at \(30^{\circ}\) above the horizontal. Find the normal force on the box in this case.

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.