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Ch 06: Work & Kinetic Energy
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 6, Problem 24a

You throw a 3.003.00-N rock vertically into the air from ground level. You observe that when it is 15.015.0 m above the ground, it is traveling at 25.025.0 m/s upward. Use the work–energy theorem to find the rock's speed just as it left the ground.

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Step 1: Start by recalling the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy. Mathematically, this is expressed as: W_{net} = rac{1}{2} m v_f^2 - rac{1}{2} m v_i^2, where v_f is the final velocity, v_i is the initial velocity, and m is the mass of the object.
Step 2: Recognize that the net work done on the rock includes both the work done by gravity and the work done by any other forces. Since no other forces are mentioned, the work done by gravity is the only contributor. The work done by gravity is given by W_{gravity} = -m g h, where g is the acceleration due to gravity (9.8 m/s²), h is the height (15.0 m), and m is the mass of the rock.
Step 3: Determine the mass of the rock using its weight. The weight of the rock is given as 3.00 N, and weight is related to mass by W = m g. Rearrange this to find m = rac{W}{g}. Substitute the given weight (3.00 N) and g = 9.8 \, \(\text{m/s}\)^2 to calculate the mass.
Step 4: Substitute the known values into the work-energy theorem. The final velocity v_f is given as 25.0 m/s, the height h is 15.0 m, and the mass m was calculated in the previous step. Rearrange the work-energy equation to solve for the initial velocity v_i: v_i = \(\sqrt{v_f^2 - \frac{2 W_{gravity}\)}{m}}.
Step 5: Perform the substitution and simplify the expression. Use the calculated mass, the given height, and the known values of g and v_f to compute v_i. This will give you the rock's speed just as it left the ground.

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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. This principle allows us to relate the forces acting on an object to its motion, providing a framework to analyze problems involving energy transfer. In this scenario, the work done by gravity and the initial kinetic energy of the rock can be used to find its speed at different heights.
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Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, calculated using the formula KE = 1/2 mv², where m is the mass and v is the velocity of the object. Understanding kinetic energy is crucial for solving the problem, as it allows us to quantify the rock's speed at various points in its trajectory and relate it to the work done against gravitational forces.
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Gravitational Potential Energy

Gravitational potential energy (PE) is the energy stored in an object due to its height above a reference point, calculated using the formula PE = mgh, where m is mass, g is the acceleration due to gravity, and h is height. In this problem, the change in potential energy as the rock rises can be equated to the work done against gravity, which is essential for determining the rock's initial speed.
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Related Practice
Textbook Question

A 1.501.50-kg book is sliding along a rough horizontal surface. At point AA it is moving at 3.213.21 m/s, and at point BB it has slowed to 1.251.25 m/s. How much work was done on the book between AA and BB?

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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 16.016.0-cm strip of the donated aorta reveal that it stretches 3.753.75 cm when a 1.501.50-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 3.003.00-N rock vertically into the air from ground level. You observe that when it is 15.015.0 m above the ground, it is traveling at 25.025.0 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 16.016.0-cm strip of the donated aorta reveal that it stretches 3.753.75 cm when a 1.501.50-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 1.141.14 cm, what is the greatest force it will be able to exert there?

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Textbook Question

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80×1061.80\(\times\)10^6 N, one 14°14° west of north and the other 14°14° east of north, as they pull the tanker 0.750.75 km toward the north. What is the total work they do on the supertanker?

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Textbook Question

Is it reasonable that a 3030-kg child could run fast enough to have 100100 J of kinetic energy?

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