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Chapter 6: Problem 77

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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Most popular questions from this chapter

A high jumper approaches the bar at \(9.0 \mathrm{~m} / \mathrm{s}\). What is the highest altitude the jumper can reach, if he does not use any additional push off the ground and is moving at \(7.0 \mathrm{~m} / \mathrm{s}\) as he goes over the bar?

The greenskeepers of golf courses use a stimpmeter to determine how "fast" their greens are. A stimpmeter is a straight aluminum bar with a V-shaped groove on which a golf ball can roll. It is designed to release the golf ball once the angle of the bar with the ground reaches a value of \(\theta=20.0^{\circ} .\) The golf ball \((\) mass \(=1.62 \mathrm{oz}=0.0459 \mathrm{~kg})\) rolls 30.0 in down the bar and then continues to roll along the green for several feet. This distance is called the "reading." The test is done on a level part of the green, and stimpmeter readings between 7 and \(12 \mathrm{ft}\) are considered acceptable. For a stimpmeter reading of \(11.1 \mathrm{ft},\) what is the coefficient of friction between the ball and the green? (The ball is rolling and not sliding, as we usually assume when considering friction, but this does not change the result in this case.)

A car of mass \(987 \mathrm{~kg}\) is traveling on a horizontal segment of a freeway with a speed of \(64.5 \mathrm{mph}\). Suddenly, the driver has to hit the brakes hard to try to avoid an accident up ahead. The car does not have an ABS (antilock braking system), and the wheels lock, causing the car to slide some distance before it is brought to a stop by the friction force between the car's tires and the road surface. The coefficient of kinetic friction is \(0.301 .\) How much mechanical energy is lost to heat in this process?

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