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Chapter 5: Problem 7

How much work do movers do (horizontally) in pushing a \(150-\mathrm{kg}\) crate \(12.3 \mathrm{~m}\) across a floor at constant speed if the coefficient of friction is \(0.70 ?\) a) 1300 J c) \(1.3 \cdot 10^{4}\) ] e) 130 ] b) 1845 J d) \(1.8 \cdot 10^{4}\) ]

Short Answer

Expert verified
The distance travelled is 12.3 meters, and the coefficient of friction between the floor and the crate is 0.70. Answer: (c) \(1.3 \cdot 10^{4} \mathrm{J}\)

Step by step solution

01

Determine the force of friction

Since the crate is moving at a constant speed, it means the force exerted by the movers is equal to the force of friction. To calculate the force of friction, we can use the following equation: _force of friction = coefficient of friction × normal force_ The normal force in this case is equal to the gravitational force acting on the crate, which can be calculated as follows: _normal force = mass × acceleration due to gravity_ Where mass is given as \(150 \mathrm{kg}\), and acceleration due to gravity can be taken as \(9.81 \mathrm{m/s^2}\). Let's calculate the normal force and then the force of friction.
02

Calculate the normal force

Using the formula for normal force, we get: _normal force = 150 \mathrm{kg} × 9.81 \mathrm{m/s^2} = 1471.5 \mathrm{N}_ Now, we have the normal force needed to calculate the force of friction.
03

Calculate the force of friction

Now, we can calculate the force of friction using the given coefficient of friction and the calculated normal force. _force of friction = 0.70 × 1471.5 \mathrm{N} = 1029.05 \mathrm{N}_ The force of friction is equal to the force exerted by the movers.
04

Calculate the work done

To calculate the work done by the movers, we can use the following equation: _work done = force × distance × cos(theta)_ Since the movers are pushing the crate horizontally, the angle (theta) would be 0 degrees, and cos(0) = 1. We have calculated the force exerted by the movers and the distance travelled is given as \(12.3 \mathrm{m}\). Now, let's plug in these values to calculate the work done.
05

Calculate the work done by the movers

Using the formula for work done, we get: _work done = 1029.05 \mathrm{N} × 12.3 \mathrm{m} × 1 = 12656.815 \mathrm{J}_ The work done by the movers is approximately \(1.3 × 10^4 \mathrm{J}\). The correct answer is (c) \(1.3 \cdot 10^{4} \mathrm{J}\).

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Most popular questions from this chapter

Which of the following is a correct unit of energy? a) \(\mathrm{kg} \mathrm{m} / \mathrm{s}^{2}\) c) \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}^{2}\) e) \(\mathrm{kg}^{2} \mathrm{~m}^{2} / \mathrm{s}^{2}\) b) \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}\) d) \(\mathrm{kg}^{2} \mathrm{~m} / \mathrm{s}^{2}\)

A small blimp is used for advertising purposes at a football game. It has a mass of \(93.5 \mathrm{~kg}\) and is attached by a towrope to a truck on the ground. The towrope makes an angle of \(53.3^{\circ}\) downward from the horizontal, and the blimp hovers at a constant height of \(19.5 \mathrm{~m}\) above the ground. The truck moves on a straight line for \(840.5 \mathrm{~m}\) on the level surface of the stadium parking lot at a constant velocity of \(8.90 \mathrm{~m} / \mathrm{s}\). If the drag coefficient \(\left(K\right.\) in \(\left.F=K v^{2}\right)\) is \(0.500 \mathrm{~kg} / \mathrm{m}\), how much work is done by the truck in pulling the blimp (assuming there is no wind)?

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An ideal spring has the spring constant \(k=440 \mathrm{~N} / \mathrm{m}\) Calculate the distance this spring must be stretched from its equilibrium position for 25 J of work to be done.
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