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Chapter 6: Problem 8

Dirk pushes on a packing box with a horizontal force of \(66.0 \mathrm{N}\) as he slides it along the floor. The average friction force acting on the box is \(4.80 \mathrm{N} .\) How much total work is done on the box in moving it $2.50 \mathrm{m}$ along the floor?

Short Answer

Expert verified
Answer: The total work done on the box is 153 J (joules).

Step by step solution

01

Define the given variables

We are given the following information: - Dirk's pushing force: \(F_{push} = 66.0 \mathrm{N}\) - Friction force: \(F_{friction} = 4.80 \mathrm{N}\) - Distance moved: \(d = 2.50 \mathrm{m}\)
02

Calculate the work done by Dirk's pushing force

The work done by a force is given by the formula \(W = F \times d\). Since Dirk's pushing force is acting in the same direction as the distance moved, we can directly use the formula: \(W_{push} = F_{push} \times d = 66.0 \mathrm{N} \times 2.50 \mathrm{m} = 165 \mathrm{J}\) (joules)
03

Calculate the work done by the friction force

The friction force is acting opposite to the direction of the distance moved. Therefore, we need to take the negative sign into account when calculating the work done: \(W_{friction} = -F_{friction} \times d = -4.80 \mathrm{N} \times 2.50 \mathrm{m} = -12.0 \mathrm{J}\)
04

Calculate the total work done on the box

The total work done is given by the sum of the work done by each force: \(W_{total} = W_{push} + W_{friction} = 165 \mathrm{J} + (-12.0 \mathrm{J}) = 153 \mathrm{J}\) The total work done on the box by Dirk's pushing force and the friction force while moving it \(2.50 \mathrm{m}\) along the floor is \(153 \mathrm{J}\).

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Most popular questions from this chapter

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A 4.0 -kg block is released from rest at the top of a frictionless plane of length \(8.0 \mathrm{m}\) that is inclined at an angle of \(15^{\circ}\) to the horizontal. A cord is attached to the block and trails along behind it. When the block reaches a point \(5.0 \mathrm{m}\) along the incline from the top, someone grasps the cord and exerts a constant tension parallel to the incline. The tension is such that the block just comes to rest when it reaches the bottom of the incline. (The person's force is a nonconservative force.) What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and the equations for constant acceleration. Which method do you prefer? A 4.0 -kg block is released from rest at the top of a frictionless plane of length \(8.0 \mathrm{m}\) that is inclined at an angle of \(15^{\circ}\) to the horizontal. A cord is attached to the block and trails along behind it. When the block reaches a point \(5.0 \mathrm{m}\) along the incline from the top, someone grasps the cord and exerts a constant tension parallel to the incline. The tension is such that the block just comes to rest when it reaches the bottom of the incline. (The person's force is a nonconservative force.) What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and the equations for constant acceleration. Which method do you prefer?
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