Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 (a) the rock's speed just as it left the ground and (b) its maximum height.

Short Answer

Expert verified
(a) The initial speed is approximately 30.3 m/s, and (b) the maximum height is 39.5 m.

Step by step solution

01

Understand the Problem

We need to find the initial speed of the rock (when it's just leaving the ground) and the maximum height it reaches using the work-energy theorem.
02

Identify Known Values

The rock's weight is 3.00 N, its velocity at 15.0 m height is 25.0 m/s, and we need to find its speed at the ground and its maximum height.
03

Calculate the Mass of the Rock

Use the gravitational force formula: \[ F = m imes g \] where \( F = 3.00 \, \text{N} \), and \( g = 9.81 \, \text{m/s}^2 \). Solve for \( m \): \[ m = \frac{3.00}{9.81} \, \text{kg} \approx 0.306 \text{ kg} \].
04

Use Work-Energy Theorem for Initial Speed

The work-energy theorem states that the change in kinetic energy is equal to the change in potential energy. Starting from ground level to 15 meters: \[ \frac{1}{2} m v^2_1 + mgh = \frac{1}{2} m v^2_2 \]where \( h = 15 \) m, \( v_2 = 25 \) m/s, and gravity \( g = 9.81 \) m/s². Solve for \( v_1 \).
05

Solve for Initial Speed \(v_1\) Equation

Input known values into the equation: \[ \frac{1}{2} m v_1^2 + m \times 9.81 \times 15 = \frac{1}{2} \times 0.306 \times 25^2 \]After calculations, solve for \( v_1 \) to find the initial speed.
06

Calculate Rock's Maximum Height

To find the maximum height, use kinetic energy at maximum height equal to zero and potential energy. \[ \frac{1}{2} m v_1^2 = mgh_{max} \]Solve for \( h_{max} \) after obtaining \( v_1 \).
07

Solve for Maximum Height Equation

After finding \( v_1 \), substitute back into \[ h_{max} = \frac{v_1^2}{2g} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy of motion. When you throw an object, like a rock, into the air, it has kinetic energy because it is moving. The formula for kinetic energy is\[ KE = \frac{1}{2}mv^2 \]where:
  • \( m \) is the mass of the object, and
  • \( v \) is its velocity.

In the exercise, as the rock is thrown upwards, it initially has a certain kinetic energy based on its speed and mass. Calculating this helps us find out the speed of the rock when it first leaves the ground. This is important because knowing the initial kinetic energy allows us to understand how much energy the rock had due to motion right from its launch point.
Potential Energy
Potential energy is stored energy due to position. For an object that is lifted against gravity, potential energy comes from height. The formula for potential energy is \[ PE = mgh \]where:
  • \( m \) is the mass,
  • \( g \) is the acceleration due to gravity \( (9.81 \, \text{m/s}^2) \), and
  • \( h \) is the height above the ground.

When the rock reaches a height of 15 m, it gains potential energy. The higher it goes, the more potential energy it stores. This shift from kinetic energy to potential energy explains how energy is conserved as the rock rises. Understanding this relationship helps in predicting the rock's future behavior, such as reaching its maximum height.
Gravitational Force
Gravitational force is a fundamental force that attracts two bodies towards each other. For objects near the Earth's surface, this force gives the object weight. The gravitational force acting on the rock can be calculated using \[ F = mg \]where:
  • \( F \) is the force or weight (in newtons),
  • \( m \) is the mass, and
  • \( g \) is the gravitational acceleration.

In this exercise, knowing the weight of the rock (3.00 N) allows us to determine its mass, important for all subsequent energy calculations. Gravitational force not only gives the rock weight but also affects how high it can go when thrown into the air. This force constantly acts as the rock moves upwards and eventually pulls it downwards, influencing both kinetic and potential energy.
Maximum Height
The maximum height of the rock is the highest point it reaches before it starts descending. At this point, all the initial kinetic energy has been converted into potential energy.
To find the maximum height, we can use the energy conversion principles illustrated in \[ \frac{1}{2} m v_1^2 = mgh_{max} \]. This means the maximum height \( h_{max} \) can be calculated by rearranging the equation to \[ h_{max} = \frac{v_1^2}{2g} \].
At maximum height, the velocity of the rock is 0 m/s because it momentarily stops before gravity pulls it back down. This concept helps us understand how energy is transferred from motion to potential due to the height increase. It's also a key component of the work-energy theorem, providing insight into how energy dictates height and motion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

While hovering, a typical flying insect applies an average force equal to twice its weight during each downward stroke. Take the mass of the insect to be 10 g, and assume the wings move an average downward distance of 1.0 cm during each stroke. Assuming 100 downward strokes per second, estimate the average power output of the insect.

(a) Suppose you cut a massless ideal spring in half. If the full spring had a force constant k, what is the force constant of each half, in terms of \(k\)? (\(Hint\): Think of the original spring as two equal halves, each producing the same force as the entire spring. Do you see why the forces must be equal?) (b) If you cut the spring into three equal segments instead, what is the force constant of each one, in terms of \(k\)?

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?

A mass \(m\) slides down a smooth inclined plane from an initial vertical height \(h\), making an angle \(\alpha\) with the horizontal. (a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height \(h\). (b) Use the work\(-\)energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if the mass had been dropped from height \(h\), independent of the angle \(\alpha\) of the incline. Explain how this speed can be independent of the slope angle. (c) Use the results of part (b) to find the speed of a rock that slides down an icy frictionless hill, starting from rest 15.0 m above the bottom.

A crate on a motorized cart starts from rest and moves with a constant eastward acceleration of \(a = 2.80 \, \mathrm{m/s}^2\). A worker assists the cart by pushing on the crate with a force that is eastward and has magnitude that depends on time according to \(F(t) = (5.40 \, \mathrm{N/s})t\). What is the instantaneous power supplied by this force at \(t = 5.00\) s?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free