/*! 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 53 A \(3.5 \mathrm{~kg}\) block is ... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(3.5 \mathrm{~kg}\) block is accelerated from rest by a compressed spring of spring constant \(640 \mathrm{~N} / \mathrm{m}\). The block leaves the spring at the spring's relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction \(\mu_{k}=0.25 .\) The frictional force stops the block in distance \(D=7.8 \mathrm{~m}\). What are (a) the increase in the thermal energy of the block-floor system, (b) the maximum kinetic energy of the block, and (c) the original compression distance of the spring?

Short Answer

Expert verified
a) 66.91 J, b) 66.91 J, c) 0.457 m

Step by step solution

01

Calculate Frictional Force

The frictional force can be determined using the formula: \[ f_k = \mu_k \cdot m \cdot g \]where \( m = 3.5 \text{ kg} \) (mass of the block), \( \mu_k = 0.25 \) (coefficient of kinetic friction), and \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity). Plug in the values: \[ f_k = 0.25 \cdot 3.5 \cdot 9.81 = 8.5775 \text{ N} \]
02

Calculate Work Done by Friction

The work done by friction, which corresponds to the increase in thermal energy, is given by: \[ W = f_k \cdot D \]where \( D = 7.8 \text{ m} \) (stopping distance). Substitute the values from Step 1: \[ W = 8.5775 \cdot 7.8 = 66.9055 \text{ J} \]Thus, the increase in thermal energy is approximately \( 66.91 \text{ J} \).
03

Use Work-Energy Principle

According to the work-energy principle, the work done on the block is equal to the change in kinetic energy. Since the block starts from rest and is brought to a stop by friction, the maximum kinetic energy \( K_{max} \) at the spring's relaxed point equals the work done by friction: \[ K_{max} = 66.9055 \text{ J} \]
04

Calculate Original Compression of the Spring

The maximum kinetic energy at the spring's relaxed length is equal to the initial elastic potential energy stored in the spring, which is calculated by: \[ K_{max} = \frac{1}{2} k x^2 \]where \( k = 640 \text{ N/m} \) (spring constant). Substitute the known values:\[ 66.9055 = \frac{1}{2} \cdot 640 \cdot x^2 \]Solve for \( x \):\[ x^2 = \frac{2 \cdot 66.9055}{640} = 0.2091 \]\[ x = \sqrt{0.2091} \approx 0.457 \text{ m} \]Thus, the original compression of the spring is about \( 0.457 \text{ m} \).

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Key Concepts

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

Kinetic Energy
Kinetic energy is a fundamental concept in physics and refers to the energy an object possesses due to its motion. It is mathematically expressed as \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.
In the context of the exercise, the block starts from rest, meaning its initial kinetic energy is zero. As the block is accelerated by the spring, its velocity increases until it reaches a maximum value when it leaves the spring. This point is when the block has its maximum kinetic energy.
The principle of conservation of energy tells us that this maximum kinetic energy is equal to the energy initially stored in the compressed spring. Once the block leaves the spring and moves across the floor, the kinetic energy is gradually converted into thermal energy due to friction until the block comes to a stop. Essentially, the kinetic energy at maximum is transformed into work done against the frictional force.
Frictional Force
Frictional force is the resistive force that opposes the motion of an object when it slides over a surface. In this exercise, it is crucial because it stops the block by transforming kinetic energy into thermal energy.
To calculate the frictional force \( f_k \), we use the formula \( f_k = \mu_k \, m \, g \), where \( \mu_k \) is the coefficient of kinetic friction, \( m \) is the mass of the block, and \( g \) is the acceleration due to gravity. This force is responsible for the conversion of the block's kinetic energy into heat as it moves over the floor.
The relationship between the frictional force and the work done can be expressed as the work-energy principle, which states that the work done by friction over a distance \( D \) is equal to the decrease in the block's kinetic energy. The calculated frictional force times the distance gives the total work done by friction, leading to the increase in the thermal energy of the block-floor system.
Elastic Potential Energy
Elastic potential energy is energy stored in elastic materials as a result of their stretching or compression. In the problem, this energy is initially stored in the compressed spring and is what accelerates the block from rest.
To find the elastic potential energy, we use the formula \( U = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the spring's relaxed length. This potential energy is entirely converted into kinetic energy as the spring returns to its relaxed length, propelling the block forward.
Solving for the compression distance is done by equating the kinetic energy at maximum, \( K_{max} \), which is observed when the block is at the spring's relaxed length, to the initial elastic potential energy. Through rearranging and solving the formula, we determine the original compression length of the spring, helping understand the spring's role in the motion of the block.

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Most popular questions from this chapter

A constant horizontal force moves a \(50 \mathrm{~kg}\) trunk \(6.0 \mathrm{~m}\) up a \(30^{\circ}\) incline at constant speed. The coefficient of kinetic friction between the trunk and the incline is \(0.20\). What are (a) the work done by the applied force and (b) the increase in the thermal energy of the trunk and incline?

The luxury liner Queen Elizabeth 2 has a diesel-electric power plant with a maximum power of \(92 \mathrm{MW}\) at a cruising speed of \(32.5\) knots. What forward force is exerted on the ship at this speed? \((1 \mathrm{knot}=1.852 \mathrm{~km} / \mathrm{h} .)\)

A spring with \(k=170 \mathrm{~N} / \mathrm{m}\) is at the top of a frictionless incline of angle \(\theta=37.0^{\circ}\). The lower end of the incline is distance \(D=1.00 \mathrm{~m}\) from the end of the spring, which is at its relaxed length. A \(2.00 \mathrm{~kg}\) canister is pushed against the spring until the spring is compressed \(0.200 \mathrm{~m}\) and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?

A block slides down an incline. As it moves from point \(A\) to point \(B\), which are \(5.0 \mathrm{~m}\) apart, force \(\vec{F}\) acts on the block, with magnitude \(2.0 \mathrm{~N}\) and directed down the incline. The magnitude of the frictional force acting on the block is \(10 \mathrm{~N}\). If the kinetic energy of the block increases by \(35 \mathrm{~J}\) between \(A\) and \(B\), how much work is done on the block by the gravitational force as the block moves from \(A\) to \(B ?\)

Ice flake is released from the edge of a hemispherical bowl whose radius \(r\) is \(22.0\) \(\mathrm{cm}\). The flake-bowl contact is frictionless. (a) How much work is done on the flake by the gravitational force during the flake's descent to the bottom of the bowl? (b) What is the change in the potential energy of the flake-Earth system during that descent? (c) If that potential energy is taken to be zero at the bottom of the bowl, what is its value when the flake is released? (d) If, instead, the potential energy is taken to be zero at the release point, what is its value when the flake reaches the bottom of the bowl? (e) If the mass of the flake were doubled, would the magnitudes of the answers to (a) through (d) increase, decrease, or remain the same?

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