A traveler pulls his 60.0 N carry-on bag a distance of 250 m along an airport floor at a constant speed with a force of 50.0 N directed at an angle of 55o above the horizontal. Find (a) the work he does, (b) the force of friction, and (c) the coefficient of kinetic friction between the bag and the floor.

To solve this problem, we need to break it down into parts and apply some fundamental physics concepts. We’ll look at the work done by the traveler, the force of friction acting on the bag, and finally, the coefficient of kinetic friction. Let’s tackle each part step by step.

Calculating the Work Done

The work done by a force is calculated using the formula:

Work (W) = Force (F) × Distance (d) × cos(θ)

In this case, the force applied by the traveler is 50.0 N, the distance is 250 m, and the angle θ is 55 degrees. We need to find the horizontal component of the force since only that contributes to the work done in the direction of motion.

First, we calculate the horizontal component of the force:

F_horizontal = F × cos(θ) = 50.0 N × cos(55°)

Using a calculator, we find:

F_horizontal ≈ 50.0 N × 0.5736 ≈ 28.68 N

Now, we can calculate the work done:

W = F_horizontal × d = 28.68 N × 250 m ≈ 7170 J

Determining the Force of Friction

Since the traveler is pulling the bag at a constant speed, the net force acting on the bag is zero. This means that the force of friction must equal the horizontal component of the pulling force:

F_friction = F_horizontal ≈ 28.68 N

Finding the Coefficient of Kinetic Friction

The force of friction can also be expressed in terms of the normal force and the coefficient of kinetic friction (μ_k):

F_friction = μ_k × F_normal

To find the normal force, we need to consider the vertical forces acting on the bag. The weight of the bag (W) is 60.0 N, and the vertical component of the pulling force (F_vertical) acts upwards:

F_vertical = F × sin(θ) = 50.0 N × sin(55°)

Calculating this gives:

F_vertical ≈ 50.0 N × 0.8192 ≈ 40.96 N

The normal force (F_normal) is the weight of the bag minus the vertical component of the pulling force:

F_normal = W - F_vertical = 60.0 N - 40.96 N ≈ 19.04 N

Now we can substitute the values into the friction equation:

28.68 N = μ_k × 19.04 N

Solving for μ_k gives:

μ_k = 28.68 N / 19.04 N ≈ 1.51

Summary of Results

• Work Done: Approximately 7170 J
• Force of Friction: Approximately 28.68 N
• Coefficient of Kinetic Friction: Approximately 1.51

This analysis shows how we can apply physics principles to real-world scenarios, like pulling a bag through an airport. Each step builds on the previous one, leading us to a comprehensive understanding of the forces at play.