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.

Ch 06: Work & Kinetic Energy

Young & Freedman Calc - University Physics 15th Edition

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Young & Freedman Calc 15th Edition

Ch 06: Work & Kinetic Energy

Problem 24a

Chapter 6, Problem 24a

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 the rock's speed just as it left the ground.

Verified step by step guidance

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.

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.

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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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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