A 6.06.0-kg box moving at 3.03.0 m/s on a horizontal, frictionless surface runs into a light spring of force constant 7575 N/cm. Use the work–energy theorem to find the maximum compression of the spring.

Ch 06: Work & Kinetic Energy

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 06: Work & Kinetic Energy

Problem 39

Chapter 6, Problem 39

A $6.0$ -kg box moving at $3.0$ m/s on a horizontal, frictionless surface runs into a light spring of force constant $75$ N/cm. Use the work–energy theorem to find the maximum compression of the spring.

Verified step by step guidance

Convert the spring constant from N/cm to N/m for consistency in SI units. Since 1 cm = 0.01 m, multiply the given spring constant (75 N/cm) by 100 to get the spring constant in N/m: $ k = 75 \times 100 = 7500 \ \text{N/m} $.

Use the work-energy theorem, which states that the work done on the box is equal to its change in kinetic energy. The box's initial kinetic energy is given by $ KE = \frac{1}{2} m v^2 $, where $ m = 6.0 \ \text{kg} $ and $ v = 3.0 \ \text{m/s} $. Substitute these values into the formula to calculate the initial kinetic energy.

At maximum compression of the spring, all the initial kinetic energy of the box is converted into elastic potential energy stored in the spring. The elastic potential energy of the spring is given by $ PE = \frac{1}{2} k x^2 $, where $ k $ is the spring constant and $ x $ is the maximum compression of the spring.

Set the initial kinetic energy of the box equal to the elastic potential energy of the spring: $ \frac{1}{2} m v^2 = \frac{1}{2} k x^2 $. Cancel out the $ \frac{1}{2} $ on both sides of the equation to simplify: $ m v^2 = k x^2 $.

Solve for $ x $, the maximum compression of the spring, by rearranging the equation: $ x = \sqrt{\frac{m v^2}{k}} $. Substitute the values $ m = 6.0 \ \text{kg} $, $ v = 3.0 \ \text{m/s} $, and $ k = 7500 \ \text{N/m} $ into the formula to find $ x $.

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Key Concepts

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

Work-Energy Theorem

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In this scenario, the kinetic energy of the box will be converted into potential energy stored in the spring as it compresses. This principle allows us to relate the initial kinetic energy of the box to the maximum compression of the spring.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, calculated using the formula KE = 0.5 * m * v^2, where m is mass and v is velocity. For the box in the question, its initial kinetic energy can be determined using its mass (6.0 kg) and velocity (3.0 m/s), which will be crucial for calculating how much energy is transferred to the spring.

Spring Potential Energy

Spring potential energy is the energy stored in a compressed or stretched spring, given by the formula PE = 0.5 * k * x^2, where k is the spring constant and x is the compression or extension from its equilibrium position. In this case, the spring constant is provided (75 N/cm), and we will use this to find the maximum compression of the spring when the kinetic energy of the box is fully converted into spring potential energy.

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Related Practice

Textbook Question

A surgeon is using material from a donated heart to repair a patient's damaged aorta and needs to know the elastic characteristics of this aortal material. Tests performed on a $16.0$ -cm strip of the donated aorta reveal that it stretches $3.75$ cm when a $1.50$ -N pull is exerted on it. What is the force constant of this strip of aortal material?

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Textbook Question

You throw a $3.00$ -N rock vertically into the air from ground level. You observe that when it is $15.0$ m above the ground, it is traveling at $25.0$ m/s upward. Use the work–energy theorem to find its maximum height.

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Textbook Question

A surgeon is using material from a donated heart to repair a patient's damaged aorta and needs to know the elastic characteristics of this aortal material. Tests performed on a $16.0$ -cm strip of the donated aorta reveal that it stretches $3.75$ cm when a $1.50$ -N pull is exerted on it. If the maximum distance it will be able to stretch when it replaces the aorta in the damaged heart is $1.14$ cm, what is the greatest force it will be able to exert there?

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Textbook Question

A $20.0$ -kg rock is sliding on a rough, horizontal surface at $8.00$ m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is $0.200$ . What average power is produced by friction as the rock stops?

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