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: 9780135159552Non è quello che usi tu?Cambia libro di testo

Tutti i libri di testo

Young & Freedman Calc 15th Edition

Ch 06: Work & Kinetic Energy

Problema 24b

Capitolo 6, Problema 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.

Guida verificata passo dopo passo

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} $.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

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.

Video consigliato:

Percorso guidato

04:10

The Work-Energy Theorem

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.

Video consigliato:

Percorso guidato

06:07

Intro to Rotational Kinetic Energy

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.

Video consigliato:

Percorso guidato

07:24

Potential Energy Graphs

Pratica correlata

Domanda del libro di testo

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.

3052

views

Domanda del libro di testo

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 $B$ ?

3384

views

Domanda del libro di testo

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?

2954

views

Domanda del libro di testo

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?

5130

views

Domanda del libro di testo

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.

7228

views

rank