/*! 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 73 A wooden block with mass 1.50 \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A wooden block with mass 1.50 \(\mathrm{kg}\) is placed against a compressed spring at the bottom of an incline of slope \(30.0^{\circ}\) (point \(A\) ). When the spring is released, it projects the block up the incline. At point \(B,\) a distance of 6.00 \(\mathrm{mup}\) the incline from \(A\) , the block is moving up the incline at 7.00 \(\mathrm{m} / \mathrm{s}\) and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_{\mathrm{k}}=0.50 .\) The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

Short Answer

Expert verified
The initial potential energy stored in the spring is approximately 119.05 J.

Step by step solution

01

Analyze the Problem

We need to determine the initial potential energy stored in the spring. This involves considering both the work done by friction and the kinetic energy of the block at point B.
02

Calculate Work Done by Friction

The work done by friction is given by the formula \( W_{friction} = -f_k d \), where \( f_k = \mu_kmg\cos(\theta) \) and \( d = 6.00 \, \text{m} \) is the distance. First, calculate the normal force: \( N = mg\cos(30^{\circ}) \). Then, \( f_k = 0.5 \times 1.50 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \cos(30^{\circ}) \approx 6.37 \, \text{N} \). Hence, \( W_{friction} = -6.37 \, \text{N} \times 6.00 \, \text{m} = -38.2 \, \text{J} \).
03

Calculate Kinetic Energy at Point B

The kinetic energy of the block at point B is calculated using \( KE_B = \frac{1}{2}mv^2 \). Substituting \( m = 1.50 \, \text{kg} \) and \( v = 7.00 \, \text{m/s} \), we find \( KE_B = 0.5 \times 1.5 \times 7^2 = 36.75 \, \text{J} \).
04

Calculate Gravitational Potential Energy at Point B

The gravitational potential energy gained by the block as it moves up the incline is \( U_B = mgh \), where \( h = d\sin(\theta) \). Thus, \( h = 6.00 \, \text{m} \times \sin(30^{\circ}) = 3.00 \, \text{m} \). Therefore, \( U_B = 1.50 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 3.00 \, \text{m} = 44.1 \, \text{J} \).
05

Apply Conservation of Energy

The initial potential energy of the spring \( PE_{spring}^{initial} \) must equal the sum of the kinetic energy at B, work done against friction, and gravitational potential energy acquired: \( PE_{spring}^{initial} = KE_B + U_B - W_{friction} \). Substituting the values: \( PE_{spring}^{initial} = 36.75 \, \text{J} + 44.1 \, \text{J} - (-38.2 \, \text{J}) \approx 119.05 \, \text{J} \).

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Key Concepts

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

Potential Energy
Potential energy is a form of energy that is stored in an object due to its position or configuration. In physics, one of the common types of potential energy is gravitational potential energy, which depends on the height of an object relative to a reference point. Another type discussed here is the elastic potential energy, stored when a spring is compressed or stretched.
  • Gravitational potential energy is calculated using the formula: \( U = mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity (usually \(9.8 \, m/s^2\)), and \( h \) is height.
  • Elastic potential energy for springs is often given by: \( PE_{spring} = \frac{1}{2}kx^2 \), where \( k \) is the spring constant, and \( x \) is the displacement from its equilibrium position.
This concept is crucial in solving problems involving springs and inclines, as it helps us understand how energy is transferred during motion. In our problem, the potential energy initially stored in the compressed spring is converted into kinetic energy and gravitational potential energy as the block moves up the incline.
Kinetic Friction
Kinetic friction occurs when two surfaces are in contact and moving relative to each other. It acts opposite to the direction of movement and works to convert kinetic energy into thermal energy, thus slowing down the moving object. The force of kinetic friction is calculated using the equation:
  • \( f_k = \mu_k N \)
  • Here, \( \mu_k \) is the coefficient of kinetic friction, which varies based on the surfaces involved.
  • \( N \) is the normal force, usually \( N = mg\cos(\theta) \) on an incline, with \( \theta \) being the angle of incline.
In the exercise, the block slides up the inclined plane, and kinetic friction does negative work. This means it removes energy from the system, which is critical when calculating the energy balance using the conservation of energy.
Conservation of Energy
The principle of conservation of energy states that in a closed system, the total energy remains constant. This means that energy cannot be created or destroyed but can be transformed from one form to another. In physics problems involving inclines and springs, this principle allows us to predict the behavior of the system by equating the sum of all forms of energy before and after an event.
  • In our problem, initially stored spring potential energy is converted into kinetic energy, gravitational potential energy, and work done against friction.
  • The equation used is \( PE_{initial} + KE_{initial} + U_{initial} = PE_{final} + KE_{final} + U_{final} \).
  • Here, initial kinetic and potential energies are zero as the block starts from rest.
  • Thus, the initial energy stored in the spring becomes the sum of kinetic energy at point B, the potential energy gained, and the work done against friction.
This allows us to solve for the unknown, in this case, the initial potential energy stored in the spring.
Incline Problems
Incline problems are a common type of physics question dealing with objects moving along sloped surfaces. These problems often involve analyzing forces like gravity, friction, and normal force, and applying principles of energy conservation. Understanding how to resolve forces along the incline is key to solving these problems.
  • The gravitational force is resolved into components parallel and perpendicular to the incline.
  • The parallel component \( mg\sin(\theta) \) causes the object to accelerate down the slope or decelerate as it moves up.
  • The normal force component \( mg\cos(\theta) \) is used to calculate friction.
  • Using these forces allows us to write equations for energy conservation and determine energy transformations such as potential energy into kinetic energy.
By mastering the vectors involved and their influences, you can solve these problems efficiently and understand the energy changes as the object moves along the incline.

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Most popular questions from this chapter

A machine part of mass \(m\) is attached to a horizontal ideal spring of force constant \(k\) that is attached to the edge of a friction-free horizontal surface. The part is pushed against the spring compressing it a distance \(x_{0},\) and then released from rest. Find the maximum (a) speed and (b) acceleration of the machine part. (c) Where in the motion do the maxima in parts (a) and (b) occur? (d) What will be the maximum extension of the spring? (e) Describe the subsequent motion of this machine part. Will it ever stop permanently?

A block with mass 0.50 \(\mathrm{kg}\) is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 \(\mathrm{m}(\mathrm{Fig} .7 .31) .\) When released, the block moves on a horizontal tabletop for 1.00 \(\mathrm{m}\) before coming to rest. The spring constant \(k\) is 100 \(\mathrm{N} / \mathrm{m}\) . What is the coefficient of kinetic friction \(\mu_{k}\) between the block and the tabletop?

A system of two paint buckets connected by a lightweight rope is released from rest with the \(12.0-\mathrm{kg}\) bucket 2.00 \(\mathrm{m}\) above the floor (Fig. 7.36\() .\) Use the principle of conservation of energy to find the speed with which this bucket strikes the floor. You can ignore friction and the mass of the pulley.

You and three friends stand at the corners of a square whose sides are 8.0 \(\mathrm{m}\) long in the middle of the gym floor, as shown in Fig. \(7.26 .\) You take your physics book and push it from one person to the other. The book has a mass of 1.5 \(\mathrm{kg}\) , and the coefficient of kinetic friction between the book and the floor is \(\mu_{\mathrm{x}}=0.25 .\) (a) The book slides from you to Beth and then from Beth to Carlos, along the lines connecting these people. What is the work done by friction during this displacement? (b) You slide the book from you to Carlos along the diagonal of the square. What is the work done by friction during this displacement? (c) You slide the book to Kim who then slides it back to you. What is the total work done by friction during this motion of the book? (d) Is the friction force on the book conservative or nonconservative? Explain.

A 1.20 \(\mathrm{kg}\) piece of cheese is placed on a vertical spring of negligible mass and force constant \(k=1800 \mathrm{N} / \mathrm{m}\) that is compressed 15.0 \(\mathrm{cm}\) . When the spring is released, how high does the cheese rise from this initial position? (The cheese and the spring are not attached.)

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