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Chapter 4: Problem 57

An \(85-\mathrm{kg}\) skier is sliding down a ski slope at a constant velocity. The slope makes an angle of \(11^{\circ}\) above the horizontal direction. (a) Ignoring any air resistance, what is the force of kinetic friction acting on the skier? (b) What is the coefficient of kinetic friction between the skis and the snow?

Short Answer

Expert verified
Answer: The force of kinetic friction acting on the skier is approximately 162.43 N, and the coefficient of kinetic friction between the skis and the snow is approximately 0.198.

Step by step solution

01

Identify the forces acting on the skier

For an \(85-\mathrm{kg}\) skier sliding down a slope with a constant velocity, there are several forces acting on them: 1. Gravity: downward force = \(mg\) where \(m=85\,\text{kg}\) and \(g=9.81\,\text{m/s}^2\). 2. The normal force: perpendicular to the slope, opposing gravity's component in that direction. 3. Kinetic friction: parallel to the slope, opposing the skier's motion. Since the skier is moving at a constant velocity, the net force acting on them is zero. Thus, the sum of the horizontal and vertical forces is zero.
02

Calculate the gravitational force component parallel to the slope

We will use the angle of \(11^{\circ}\) to break down the gravitational force into components parallel and perpendicular to the slope. The gravitational force parallel to the slope is given by: \(F_{parallel} = mg\sin{\theta}\) where \(m = 85\,\text{kg}\), \(g = 9.81\,\text{m/s}^2\), and \(\theta = 11^{\circ}\). Substituting the values, we get: \(F_{parallel} = 85*9.81*\sin(11^{\circ}) \approx 162.43\,\text{N}\)
03

Determine the force of kinetic friction

Since the skier is moving at a constant velocity, the force of kinetic friction is equal in magnitude and opposite in direction to the gravitational force component parallel to the slope. Therefore, the force of kinetic friction, \(f_{k}\), is: \(f_{k} = F_{parallel} \approx 162.43\,\text{N}\)
04

Calculate the normal force acting on the skier

The normal force is equal but opposite to the gravitational force component perpendicular to the slope. The gravitational force perpendicular to the slope is given by: \(F_{perpendicular} = mg\cos{\theta}\) Substituting the values, we get: \(F_{perpendicular} = 85*9.81*\cos(11^{\circ}) \approx 820.61\,\text{N}\) Since these two forces are in equilibrium, the normal force, \(N\), equals \(F_{perpendicular} \approx 820.61\,\text{N}\)
05

Calculate the coefficient of kinetic friction

We use the following expression to find the coefficient of kinetic friction, \(\mu_k\): \(\mu_k = \frac{f_{k}}{N}\) Substituting the values we found in the previous steps, we get: \(\mu_k = \frac{162.43}{820.61} \approx 0.198\) So, the force of kinetic friction acting on the skier is approximately \(162.43\,\text{N}\), and the coefficient of kinetic friction between the skis and the snow is approximately \(0.198\).

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