You throw a 3.003.003.00-N rock vertically into the air from ground level. You observe that when it is 15.015.015.0 m above the ground, it is traveling at 25.025.025.0 m/s upward. Use the work–energy theorem to find its maximum height.

Ch 06: Work & Kinetic Energy

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

All textbooks

Young & Freedman Calc 15th Edition

Ch 06: Work & Kinetic Energy

Problem 24b

Chapter 6, Problem 24b

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.

Verified step by step guidance

Step 1: Start by identifying the given values and the work-energy theorem. The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy. Here, the rock has a weight of 3.00 N, an initial velocity of 25.0 m/s at a height of 15.0 m, and we need to find its maximum height.

Step 2: Calculate the mass of the rock using its weight and the acceleration due to gravity. The formula is $ m = \frac{W}{g} $, where $ W $ is the weight (3.00 N) and $ g $ is the acceleration due to gravity (9.8 m/s²).

Step 3: Use the work-energy theorem to relate the kinetic energy and potential energy at the given height (15.0 m) to the maximum height. The total mechanical energy at any point is conserved, so $ KE_{initial} + PE_{initial} = PE_{max} $. Kinetic energy is given by $ KE = \frac{1}{2}mv^2 $, and potential energy is given by $ PE = mgh $.

Step 4: Substitute the known values into the equations. At 15.0 m, calculate the kinetic energy using $ KE = \frac{1}{2}mv^2 $ and the potential energy using $ PE = mgh $. Add these to find the total mechanical energy.

Step 5: At the maximum height, the rock's velocity is zero, so all the mechanical energy is converted into potential energy. Use $ PE_{max} = mgh_{max} $ and solve for $ h_{max} $ by rearranging the equation to $ h_{max} = \frac{E_{total}}{mg} $.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

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 context, the work done by the gravitational force as the rock ascends can be calculated, and this work will equal the difference between the kinetic energy at the height of 15.0 m and the kinetic energy at the maximum height.

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. In this problem, the rock's kinetic energy at 15.0 m can be determined using its speed of 25.0 m/s, which will help in applying the work-energy theorem to find the maximum height.

Potential Energy

Potential energy, particularly gravitational potential energy, is the energy stored in an object due to its height above a reference point, calculated as PE = m * g * h, where g is the acceleration due to gravity and h is the height. As the rock rises, its potential energy increases, and at its maximum height, all kinetic energy will have converted into potential energy.

Recommended video:

Guided course

07:24

Potential Energy Graphs

Related Practice

Textbook Question

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.

views

Textbook Question

A $1.50$ -kg book is sliding along a rough horizontal surface. At point $A$ it is moving at $3.21$ m/s, and at point $B$ it has slowed to $1.25$ m/s. How much work was done on the book between $A$ and $B$ ?

views

Textbook Question

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of $1.80 \times 10^{6}$ N, one $14 degrees$ west of north and the other $14 degrees$ east of north, as they pull the tanker $0.75$ km toward the north. What is the total work they do on the supertanker?

views

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?

views

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

views