/*! 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 77 You have decided to move a refri... [FREE SOLUTION] | 91Ó°ÊÓ

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You have decided to move a refrigerator (mass \(=81.3 \mathrm{~kg}\), including all the contents) to the other side of the room. You slide it across the floor on a straight path of length \(6.35 \mathrm{~m}\), and the coefficient of kinetic friction between floor and fridge is \(0.437 .\) Happy about your accomplishment, you leave the apartment. Your roommate comes home, wonders why the fridge is on the other side of the room, picks it up (you have a strong roommate!), carries it back to where it was originally, and puts it down. How much net mechanical work have the two of you done together?

Short Answer

Expert verified
Answer: The net mechanical work done by both the person and the roommate together is \(7264.29 \mathrm{~J}\).

Step by step solution

01

Identify the known values

We know the following values: - Mass of the refrigerator: \(m = 81.3 \mathrm{~kg}\) - Length of the path: \(d = 6.35 \mathrm{~m}\) - Coefficient of kinetic friction: \(\mu_k = 0.437\)
02

Calculate the force of friction

To calculate the force of friction, we first need to determine the gravitational force acting on the fridge. This is given by \(F_g = m \times g\) where \(g\) is the gravitational acceleration, which is approximately \(9.8 \mathrm{~m/s^2}\). So, \(F_g = 81.3 \mathrm{~kg} \times 9.8 \mathrm{~m/s^2} = 796.74 \mathrm{~N}\) The force of friction is given by \(F_f = \mu_k \times F_g\) So, \(F_f = 0.437 \times 796.74 \mathrm{~N} = 347.73 \mathrm{~N}\)
03

Calculate the work done while sliding the fridge

The work done against friction in sliding the fridge is given by \(W_1 = F_f \times d\) So, \(W_1 = 347.73 \mathrm{~N} \times 6.35 \mathrm{~m} = 2206.49 \mathrm{~J}\) (Joules)
04

Calculate the work done while lifting and carrying the fridge back

The work done against gravity in lifting the fridge is given by \(W_2 = F_g \times d\) So, \(W_2 = 796.74 \mathrm{~N} \times 6.35 \mathrm{~m} = 5057.80 \mathrm{~J}\) (Joules)
05

Calculate the net mechanical work

The net mechanical work done by both the person and the roommate is the sum of the work done by each of them: \(W_{net} = W_1 + W_2\) So, \(W_{net} = 2206.49 \mathrm{~J} + 5057.80 \mathrm{~J} = 7264.29 \mathrm{~J}\) (Joules) The net mechanical work done by both the person and the roommate together is \(7264.29 \mathrm{~J}\).

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Key Concepts

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

Mechanical Work
When moving objects, understanding mechanical work can help us determine the effort required to perform a task. Mechanical work is defined as the amount of energy transferred when a force is applied over a distance. It is calculated using the formula: \[ W = F \times d \]where:
  • \( W \) is the work done, measured in Joules (J)
  • \( F \) is the force applied, in Newtons (N)
  • \( d \) is the distance over which the force is applied, in meters (m)
In our example, after calculating the force of friction and gravitational force, we use these forces to find the mechanical work done during different stages of moving the refrigerator. Whether you slide it across the floor or lift it up, the work done indicates how much energy was spent in overcoming these forces. Every time a force overcomes a resistance like gravity or friction, mechanical work is being performed.
Force of Friction
The force of friction is an opposing force that acts against the motion of two surfaces sliding past each other. It is crucial in many real-world applications, like moving a fridge across a floor. In our problem, the force of friction can be calculated using the equation:\[ F_f = \mu_k \times F_g \]where:
  • \( F_f \) is the force of friction, in Newtons (N)
  • \( \mu_k \) is the coefficient of kinetic friction, a unitless value that describes the friction level between two surfaces
  • \( F_g \) is the gravitational force, also in Newtons (N)
Kinetic friction, specifically, pertains to objects in motion—unlike static friction, which comes into play when objects are at rest. This force affects how much work is needed to move objects, as more friction requires more energy to overcome. Calculating the force of friction helps determine the total effort needed, seen in the amount of work done to slide the refrigerator across the room.
Gravitational Force
Gravitational force is a fundamental force acting vertically downwards towards the center of the Earth, affecting all objects with mass. It is calculated by multiplying an object's mass by the acceleration due to gravity: \[ F_g = m \times g \]where:
  • \( F_g \) is the gravitational force, in Newtons (N)
  • \( m \) is the mass of the object, in kilograms (kg)
  • \( g \) is the acceleration due to gravity, typically \(9.8 \ m/s^2\) on Earth's surface
In the context of lifting the refrigerator, the gravitational force is the force that needs to be overcome to elevate the object against gravity. This requires significant energy, reflecting in the mechanical work done during lifting. The interaction between gravitational forces and the work done is essential in understanding how various forces impact energy expenditure while moving objects vertically or horizontally.

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Most popular questions from this chapter

An arrow is placed on a bow, the bowstring is pulled back, and the arrow is shot straight up into the air; the arrow then comes back down and sticks into the ground. Describe all of the changes in work and energy that occur.

A 70.0 -kg skier moving horizontally at \(4.50 \mathrm{~m} / \mathrm{s}\) encounters a \(20.0^{\circ}\) incline. a) How far up the incline will the skier move before she momentarily stops, ignoring friction? b) How far up the incline will the skier move if the coefficient of kinetic friction between the skies and snow is \(0.100 ?\)

A spring with \(k=10.0 \mathrm{~N} / \mathrm{cm}\) is initially stretched \(1.00 \mathrm{~cm}\) from its equilibrium length. a) How much more energy is needed to further stretch the spring to \(5.00 \mathrm{~cm}\) beyond its equilibrium length? b) From this new position, how much energy is needed to compress the spring to \(5.00 \mathrm{~cm}\) shorter than its equilibrium position?

A 1.00 -kg block is resting against a light, compressed spring at the bottom of a rough plane inclined at an angle of \(30.0^{\circ}\); the coefficient of kinetic friction between block and plane is \(\mu_{\mathrm{k}}=0.100 .\) Suppose the spring is compressed \(10.0 \mathrm{~cm}\) from its equilibrium length. The spring is then released, and the block separates from the spring and slides up the incline a distance of only \(2.00 \mathrm{~cm}\) beyond the spring's normal length before it stops. Determine a) the change in total mechanical energy of the system and b) the spring constant \(k\).

A 0.100 -kg ball is dropped from a height of \(1.00 \mathrm{~m}\) and lands on a light (approximately massless) cup mounted on top of a light, vertical spring initially at its equilibrium position. The maximum compression of the spring is to be \(10.0 \mathrm{~cm}\). a) What is the required spring constant of the spring? b) Suppose you ignore the change in the gravitational energy of the ball during the 10 -cm compression. What is the percentage difference between the calculated spring constant for this case and the answer obtained in part (a)?

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