/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 A small box with mass \(0.600 \m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A small box with mass \(0.600 \mathrm{~kg}\) is placed against a compressed spring at the bottom of an incline that slopes upward at \(37.0^{\circ}\) above the horizontal. The other end of the spring is attached to a wall. The coefficient of kinetic friction between the box and the surface of the incline is \(\mu_{k}=0.400\) The spring is released and the box travels up the incline, leaving the spring behind. What minimum elastic potential energy must be stored initially in the spring if the box is to travel \(2.00 \mathrm{~m}\) from its initial position to the top of the incline?

Short Answer

Expert verified
The calculation made in Step 2 provides the minimal potential energy stored in the spring. This implies that to cover the specified distance uphill overcoming friction and gravity, the spring must initially store this amount of potential energy.

Step by step solution

01

Identify the forces acting on the box

Firstly, two main forces are acting on the box as it moves up the incline, namely:1. The gravitational force: \( mg \cos(37.0^{\circ})\)2. The frictional force: \( \mu_{k} mg \cos(37.0^{\circ})\)
02

Calculate the work done against these forces

The work done against both the gravitational and the frictional forces can be calculated by the formula: \[ d = 2.00 m \]\[ W = d ( F_{g} + F_{f} ) \]To calculate for the work, input the given values into the equation:\[ W = 2.00 m \times ((0.600 kg \times 9.8 m/s^2 \times \cos(37.0^{\circ})) + (0.400 \times 0.600 kg \times 9.8 m/s^2 \times \cos(37.0^{\circ}))) \]
03

Calculate the elastic potential energy stored in the spring

The work done on the box by these forces is equal to the minimum elastic potential energy that must initially be stored in the spring for the box to be able to travel to the top of the incline. After performing the above calculation for work done, you have the answer for the stored potential energy.

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 91Ó°ÊÓ!

Key Concepts

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

Kinetic Friction
Imagine sliding a heavy box across the floor; the force you feel resisting the motion is kinetic friction. In our exercise, kinetic friction plays a crucial role as it opposes the movement of the box up the incline. It's the force that acts between moving surfaces and is calculated with the formula \( F_{k} = \mu_{k} \times N \), where \( \mu_{k} \) is the coefficient of kinetic friction and \( N \) is the normal force—in this case, \( mg \cos(\theta) \). Since the incline angle and mass are known, and gravity \( (g) \) is a constant \( 9.8 m/s^2 \), we can determine the force of kinetic friction accurately.

This force does work against the box, consuming some of the energy provided by the spring. We assess the energy lost to friction by multiplying the force of kinetic friction by the distance the box travels, demonstrating how essential it is to consider friction when calculating motion on surfaces.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics. It tells us that work done on an object is equal to the change in its energy. In the context of our exercise, the work done by the spring on the box is transformed into kinetic energy as the box begins to move, and some of this energy is then converted into elastic potential energy due to the box climbing the incline.

The work done against gravity and friction is given by the equation \( W = Fd \), where \( F \) is the total force working against the movement, and \( d \) is the distance the box travels. By calculating the work done against gravitational and frictional forces, we can determine the minimum elastic potential energy needed in the spring for the box to reach the specified distance on the incline. This illustrates how the work-energy principle links the stored energy in the spring to the mechanical work required to overcome opposing forces.
Gravitational Force
Every object on Earth experiences the pull of gravity, a force that draws objects toward the center of the Earth. The gravitational force on the box in our exercise is central to solving the problem. It is this force that the box must work against to travel up the incline.

Gravitational force is determined by the mass of the object and the acceleration due to gravity (\( g = 9.8 m/s^2 \) on Earth). On an incline, the effective gravitational force the box experiences is \( mg \cos(\theta) \) whereas \( \theta \) is the angle of the incline, and \( \cos(\theta) \) represents the fraction of the gravitational force acting parallel to the surface of the incline. Gravitational force contributes to the total work that needs to be done by the elastic potential energy stored in the spring, allowing us to understand how much energy is necessary for the box to ascend the given distance on the incline.

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

A 25.0 kg child plays on a swing having support ropes that are 2.20 m long. Her brother pulls her back until the ropes are 42.0° from the vertical and releases her from rest. (a) What is her potential energy just as she is released, compared with the potential energy at the bottom of the swing’s motion? (b) How fast will she be moving at the bottom? (c) How much work does the tension in the ropes do as she swings from the initial position to the bottom of the motion?

Two blocks are attached to either end of a light rope that passes over a light, frictionless pulley suspended from the ceiling. One block has mass \(8.00 \mathrm{~kg},\) and the other has mass \(6.00 \mathrm{~kg}\). The blocks are released from rest. (a) For a \(0.200 \mathrm{~m}\) downward displacement of the \(8.00 \mathrm{~kg}\) block, what is the change in the gravitational potential energy associated with each block? (b) If the tension in the rope is \(T\), how much work is done on each block by the rope? (c) Apply conservation of energy to the system that includes both blocks. During the \(0.200 \mathrm{~m}\) downward displacement, what is the total work done on the system by the tension in the rope? What is the change in gravitational potcntial energy associated with the system? Use energy conservation to find the speed of the \(8.00 \mathrm{~kg}\) block after it has descended \(0.200 \mathrm{~m} .\)

The maximum height a typical human can jump from a crouched start is about 60 cm. By how much does the gravitational potential energy increase for a 72 kg person in such a jump? Where does this energy come from?

A block with mass \(m=\) \(\begin{array}{lll}0.200 \mathrm{~kg} & \text { is placed against a com- }\end{array}\) pressed spring at the bottom of a ramp that is at an angle of \(53.0^{\circ}\) above the horizontal. The spring has \(8.00 \mathrm{~J}\) of elastic potential energy stored in it. The spring is released, and the block moves up the incline. After the block has traveled a distance of \(3.00 \mathrm{~m},\) its speed is \(4.00 \mathrm{~m} / \mathrm{s} .\) What is the magnitude of the friction force that the ramp exerts on the block while the block is moving?

In one day, a 75 kg mountain climber ascends from the 1500 m level on a vertical cliff to the top at 2400 m. The next day, she descends from the top to the base of the cliff, which is at an elevation of 1350 m. What is her change in gravitational potential energy (a) on the first day and (b) on the second day?

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.