/*! 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 19 A 750.0 -kg boulder is raised fr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 750.0 -kg boulder is raised from a quarry 125 \(\mathrm{m}\) deep by a long uniform chain having a mass of 575 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 to be lifted out at maximum acceleration if it started from rest?

Short Answer

Expert verified
(a) Maximum acceleration is 8.70 m/s². (b) It takes approximately 5.34 seconds.

Step by step solution

01

Calculate the Weight of the Boulder and Chain

First, we need to calculate the total weight of the boulder and the chain. The weight of the boulder is given by\( W_b = m_b \cdot g = 750.0 \times 9.81 \),where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.The weight of the chain is similarly given by\( W_c = m_c \cdot g = 575 \times 9.81 \).
02

Determine Maximum Total Tension the Chain Can Support

The chain can support a maximum tension 2.50 times its own weight. Therefore, the maximum tension that the chain can support is:\( T_{max} = 2.50 \times W_c = 2.50 \times (575 \times 9.81) \).
03

Calculate the Maximum Net Force Available for Accelerating the Boulder

The maximum tension must support both the boulder and accelerate it out of the quarry. The net force for acceleration \( F_{net} \) is given by:\( F_{net} = T_{max} - W_b \).
04

Find the Maximum Acceleration of the Boulder

Now, use Newton's second law to find the maximum acceleration \( a_{max} \).\( F_{net} = m_b \cdot a_{max} \),so,\( a_{max} = \frac{F_{net}}{m_b} \).
05

Calculate the Time Taken to Lift the Boulder

Since the boulder starts from rest, the equation for motion is\( d = \frac{1}{2} a_{max} t^2 \),where \( d = 125 \, \text{m} \) is the depth of the quarry. Solving for \( t \), we get:\( t = \sqrt{\frac{2d}{a_{max}}} \).

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.

Understanding Tension
In mechanics, *tension* refers to the force transmitted through a string, cable, or chain when it is pulled tight by forces acting from opposite ends. It is a key concept in this exercise as the chain raises the boulder from the quarry. - The tension within the chain must counteract the weight of both the boulder and itself.
- The chain can withstand a tension up to 2.50 times its own weight before breaking.
To calculate the tension, you start with the weight of the chain. Since the maximum tension is a multiplier of the chain's weight, it's crucial to know the chain's mass, which is 575 kg in this case. The force of gravity acting on this mass is also a factor, using the formula: \[ T = 2.50 imes (m_c imes g) \] where \( m_c \) is the mass of the chain and \( g = 9.81 \, \text{m/s}^2 \) is the gravitational acceleration.
Decoding Acceleration
*Acceleration* indicates how fast an object can change its velocity. In the context of the boulder being lifted, it's how quickly it speeds up as it ascends. - Acceleration is directly linked to the net force acting upon an object.
- It is determined by rearranging Newton's second law to solve for acceleration.
When lifting the boulder, the chain's tension that goes into accelerating it is: \[ F_{net} = T_{max} - W_b \] where \( W_b \) is the weight of the boulder. The acceleration \( a_{max} \) can then be obtained using: \[ a_{max} = \frac{F_{net}}{m_b} \] This formula showcases that the greater the available net force from the tension, the higher the possible acceleration.
Explaining Newton's Second Law
Newton's second law is fundamental in understanding how objects behave when forces are applied. It is stated as \( F = ma \), where \( F \) is the net force applied to an object, \( m \) is the mass, and \( a \) is the acceleration. - The boulder's acceleration depends on the net force derived from the chain's tension.
- If the tension exceeds the weight of the boulder, the excess force accelerates it upward.
Newton's second law allows us to calculate this acceleration by knowing the net force acting on the boulder: \[ a_{max} = \frac{F_{net}}{m_b} \] This simplicity shows the power of Newton's second law. It connects mass, acceleration, and force, serving as a predictive tool to understand how the boulder reacts to tension.
Calculating Weight and its Role
*Weight* is a key factor in mechanics, representing the force due to gravity acting on a mass. The weight of both the boulder and the chain affects the maximum tension required for lifting. - Weight is calculated by multiplying mass by gravitational acceleration: \( W = m \times g \).
- For both the boulder and the chain, the individual weights are needed to sum the forces that the chain must hold.
In this exercise: - Boulder weight: \( W_b = 750.0 \times 9.81 \approx 7357.5 \, \text{N} \) - Chain weight: \( W_c = 575 \times 9.81 \approx 5641.75 \, \text{N} \) These weights ensure that the total force does not exceed the chain's maximum tension, preventing the chain from breaking and allowing proper movement of the boulder.

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 physics major is working to pay his college tuition by performing in a traveling carnival. He rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, he travels in a vertical circle with a radius of 13.0 \(\mathrm{m} .\) The physics major has mass \(70.0 \mathrm{kg},\) and his motorcycle has mass 40.0 \(\mathrm{kg}\) . (a) What minimum speed must he have at the top of the circle if the tires of the motorcycle are not to lose contact with the sphere? (b) At the bottom of the circle, his speed is twice the value calculated in part (a). What is the magnitude of the normal force exerted on the motorcycle by the sphere at this point?

You are part of a design team for future exploration of the planet Mars, where \(g=3.7 \mathrm{m} / \mathrm{s}^{2} .\) An explorer is to step out of a survey vehicle traveling horizontally at 33 \(\mathrm{m} / \mathrm{s}\) when it is 1200 \(\mathrm{m}\) above the surface and then freely for 20 \(\mathrm{s}\) . At that time, a portable advanced propulsion system (PAPS) is to exert a constant force that will decrease the explorer's speed to zero at the instant she touches the surface. The total mass (explorer, suit, equipment, and PAPS) is 150 \(\mathrm{kg} .\) Assume the change in mass of the PAPS to be negligible. Find the horizontal and vertical components of the force the PAPS must exert, and for what interval of time the PAPS must exert it. You can ignore air resistance.

A box with weight \(w\) is pulled at constant speed along a level floor by a force \(\vec{\boldsymbol{F}}\) that is at an angle \(\theta\) above the horizontal. The coefficient of kinetic friction between the floor and box is \(\mu_{\mathrm{k}}\) (a) In terms of \(\theta, \mu_{\mathrm{k}},\) and \(w\) calculate \(F .\) (b) For \(w=400 \mathrm{N}\) and \(\mu_{\mathrm{k}}=0.25,\) calculate \(F\) for \(\theta\) ranging from \(0^{\circ}\) to \(90^{\circ}\) in increments of \(10^{\circ} .\) Graph \(F\) versus \(\theta\) .(c) From the general expression in part (a), calculate the value of \(\theta\) for which the value of \(F\) , required to maintain constant speed, is a minimum. (Hint: At a point where a function is minimum, what are the first and second derivatives of the function? Here \(F\) is a function of \(\theta . )\) For the special case of \(w=400 \mathrm{N}\) and \(\mu_{\mathrm{k}}=0.25\) evaluate this optimal \(\theta\) and compare your result to the graph you constructed in part (b).

Block \(A\) in Fig. P5.72 weighs 60.0 \(\mathrm{N} .\) The coefficient of static friction between the block and the surface on which it rests is \(0.25 .\) The weight \(w\) is 12.0 \(\mathrm{N}\) and the system is in equilibrium. (a) Find the friction force exerted on block \(A .\) (b) Find the maximum weight \(w\) for which the system will remain in equilibrium.

You throw a baseball straight up. The drag force is proportional to \(v^{2} .\) In terms of \(g,\) what is the \(y\) -component of the ball's acceleration when its speed is half its terminal speed and (a) it is moving up? It is moving back down?

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.