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Ch 05: Applying Newton's Laws
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 05: Applying Newton's LawsProblem 19b
Chapter 5, Problem 19b

A 750.0750.0-kg boulder is raised from a quarry 125125 m deep by a long uniform chain having a mass of 575575 kg. This chain is of uniform strength, but at any point it can support a maximum tension no greater than 2.502.50 times its weight without breaking. How long does it take to be lifted out at maximum acceleration if it started from rest?

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Determine the total weight of the boulder and the chain. The weight of the boulder is given by \( W_b = m_b g \), where \( m_b = 750.0 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). The weight of the chain is \( W_c = m_c g \), where \( m_c = 575 \, \text{kg} \).
Calculate the maximum tension the chain can support. The chain can support a maximum tension of \( T_{max} = 2.50 \times W_c \). This tension must account for the combined weight of the boulder and the chain, as well as the force required for acceleration.
Set up the force equation for the system. The total force required to lift the boulder and chain with acceleration \( a \) is \( T = (m_b + m_c)(g + a) \). Ensure that \( T \leq T_{max} \) to avoid breaking the chain.
Solve for the maximum acceleration \( a \) that satisfies \( T \leq T_{max} \). Rearrange the inequality \( (m_b + m_c)(g + a) \leq T_{max} \) to isolate \( a \): \( a \leq \frac{T_{max}}{m_b + m_c} - g \).
Use the kinematic equation to find the time required to lift the boulder out of the quarry. The equation \( d = \frac{1}{2} a t^2 \) relates the depth \( d = 125 \, \text{m} \), the acceleration \( a \), and the time \( t \). Solve for \( t \): \( t = \sqrt{\frac{2d}{a}} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This principle is crucial for understanding how forces affect the motion of the boulder and the chain. In this scenario, the net force will determine the maximum acceleration at which the boulder can be lifted, factoring in both gravitational and tension forces.
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Tension in a Rope or Chain

Tension is the force transmitted through a rope or chain when it is pulled tight by forces acting from opposite ends. In this problem, the chain must support the weight of the boulder while also providing the necessary force to accelerate it upwards. The maximum tension that the chain can withstand is a critical factor in determining how quickly the boulder can be lifted without breaking the chain.
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Kinematic Equations of Motion

Kinematic equations describe the motion of objects under constant acceleration. These equations relate displacement, initial velocity, final velocity, acceleration, and time. In this case, they will be used to calculate the time it takes for the boulder to be lifted from rest to a certain height under maximum acceleration, providing a mathematical framework to solve the problem.
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Related Practice
Textbook Question

A light rope is attached to a block with mass 4.004.00 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass mm is suspended from the other end. When the blocks are released, the tension in the rope is 15.015.0 N. What is the acceleration of either block?

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Textbook Question

When jumping straight up from a crouched position, an average person can reach a maximum height of about 6060 cm. During the jump, the person's body from the knees up typically rises a distance of around 5050 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. Draw a free-body diagram of the person during the jump.

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Textbook Question

A light rope is attached to a block with mass 4.004.00 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 15.015.0 N. Find mm.

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Textbook Question

When jumping straight up from a crouched position, an average person can reach a maximum height of about 6060 cm. During the jump, the person's body from the knees up typically rises a distance of around 5050 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. With what initial speed does the person leave the ground to reach a height of 6060 cm?

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Textbook Question

A 550550-N physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plus elevator is 850850 kg. As the elevator starts moving, the scale reads 450450 N. Find the acceleration of the elevator (magnitude and direction).

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Textbook Question

A light rope is attached to a block with mass 4.004.00 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass mm is suspended from the other end. When the blocks are released, the tension in the rope is 15.015.0 N. How does the tension compare to the weight of the hanging block?

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