/*! 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 42 Worker Pushes Block A worker pus... [FREE SOLUTION] | 91Ó°ÊÓ

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Worker Pushes Block A worker pushed a \(27 \mathrm{~kg}\) block \(9.2 \mathrm{~m}\) along a level floor at constant speed with a force directed \(32^{\circ}\) below the horizontal. If the coefficient of kinetic friction between block and floor was \(0.20\), what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?

Short Answer

Expert verified
Work done by the worker’s force is approximately 518.5 J. The increase in thermal energy is 486.36 J.

Step by step solution

01

- Identify the forces

The worker’s force, kinetic friction force, and gravitational force act on the block. Analyze the vertical and horizontal components.
02

- Calculate Normal Force

Sum the vertical forces to find the normal force: \[ N = mg + F \, \text{sin}(32^{\circ}) \] where \( N \) is the normal force, \( m = 27 \) kg is the mass, \( g = 9.8 \) m/s\textsuperscript{2} is the gravitational acceleration, and \( F \) is the force applied by the worker.
03

- Kinetic Friction Force

Determine the kinetic friction force, \[ f_k = \mu_k N \] \( \mu_k = 0.20 \) is the coefficient of kinetic friction.
04

- Horizontal forces and solving for F

Set up the equilibrium condition for horizontal forces: \[ F \, \text{cos}(32^{\circ}) - f_k = 0 \] and solve for \( F \).
05

- Work Done by the Worker’s Force

Calculate the work done, \[ W = F d \, \text{cos}(32^{\circ}) \] where \( d = 9.2 \) m is the distance.
06

- Increase in Thermal Energy

The increase in thermal energy equals the work done against friction: \[ \Delta E_{thermal} = f_k d \]

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Key Concepts

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

Newton's laws of motion
Understanding Newton's laws of motion is crucial for tackling this problem. Newton's First Law states that an object will remain at rest or in uniform motion unless acted upon by a net external force. In this exercise, the block moves at a constant speed, which means the net force acting on it is zero. Newton's Second Law tells us that the sum of the forces acting on an object is equal to the object’s mass times its acceleration. Because the block moves at a constant speed, its acceleration is zero, meaning the forces are balanced.

In the vertical direction, the normal force and the vertical component of the worker's force counteract the gravitational force. Horizontally, the worker's applied force balances the kinetic friction force. Newton's laws guide us through setting up the equilibrium conditions needed to solve for unknowns.
Work-energy principle
The work-energy principle is vital to decomposing what happens when the worker pushes the block. It states that the work done by all forces on an object results in a change in its kinetic energy. Since the block moves at constant speed, its kinetic energy does not change. Therefore, the work done by the worker must go into overcoming the kinetic friction and transforming into thermal energy.

This means:
  • The work done by the worker is calculated using the force applied and the distance moved.
  • In our case, the worker's force is partially vertical, so we consider only the horizontal component for calculating work done.
Using these principles, we break down the forces and calculate the work done, which directly correlates to the increase in thermal energy.
Kinetic friction force
Friction is a resistive force that acts opposite to the direction of motion. Kinetic friction occurs when two objects slide past each other. The magnitude of kinetic friction force can be calculated using: \ f_k = \mu_k N where \( \mu_k = 0.20 \) is the coefficient of kinetic friction and \( N \) is the normal force.

In this exercise, kinetic friction plays a key role because the worker's force must overcome this resistive force to keep the block moving at a constant speed. By calculating the normal force and then using it to find the kinetic friction, we ensure that the force applied by the worker counterbalances the resistance due to friction.
Normal force calculation
The normal force is the force exerted by a surface to support the weight of an object resting on it. Its direction is perpendicular to the surface. For the block in the given problem, the normal force is affected by both the weight of the block and the vertical component of the worker’s applied force.

The normal force can be found using the formula: \ N = mg + \text{F sin}(32^{\circ}) where \( m = 27 \) kg, \( g = 9.8 \) m/s\( ^2 \) and \( F \) is the applied force. By summing up these forces, we accurately determine the normal force that facilitates calculating the kinetic friction force.
Thermal energy
Thermal energy in this context refers to the internal energy generated due to the friction between the block and the floor. When the block is pushed, kinetic friction converts part of the worker's applied force into heat, increasing the thermal energy of the block-floor system.

The increase in thermal energy can be calculated as: \ \text{\Delta E_{thermal}} = f_k d where \( f_k \) is the kinetic friction force and \( d = 9.2 \) m is the distance moved. This demonstrates the energy transformation from mechanical work done by the worker to heat energy due to friction, underscoring the link between physical exertion and thermal dynamics in the block-floor system.

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