Question

A 70.0 -kg skier moving horizontally at \(4.50 \mathrm{~m} / \mathrm{s}\) encounters a \(20.0^{\circ}\) incline. a) How far up the incline will the skier move before she momentarily stops, ignoring friction? b) How far up the incline will the skier move if the coefficient of kinetic friction between the skies and snow is \(0.100 ?\)

Short answer

a) When ignoring friction, the skier travels 2.989 m up the incline before momentarily stopping. b) When taking friction into account, the skier travels 4.218 m up the incline before momentarily stopping.

Step-by-step solution

a) Distance up the incline without friction

First, we will calculate the skier's initial kinetic energy (KE) using the formula KE = (1/2)mv^2, where m is the mass (70 kg) and v is the initial velocity (4.50 m/s): KE = (1/2)(70 kg)(4.50 m/s)^2 = 709.125 J Now, we will find the potential energy at the highest point on the incline, where the skier momentarily stops and her kinetic energy is zero. We will use the formula PE = mgh, where h is the height reached, and g is the acceleration due to gravity (9.81 m/s^2). We will equate this potential energy with the initial kinetic energy: 709.125 J = (70 kg)(9.81 m/s^2)h Let's solve for h: h = 709.125 J / [(70 kg)(9.81 m/s^2)] = 1.024 m Now, we will find the distance up the incline, d, using the equation d = h/sin(angle), where angle is the incline angle (20°): d = 1.024 m / sin(20°) = 2.989 m So, the skier will move 2.989 m up the incline before she momentarily stops, ignoring friction.

b) Distance up the incline with friction

Now, we will consider the effects of friction. The net force acting on the skier parallel to the incline is given by the equation F_net = F_gravity + F_friction, where F_gravity is the force due to gravity acting parallel to the incline, and F_friction is the force due to friction. F_gravity = mg * sin(angle) = (70 kg)(9.81 m/s^2) * sin(20°) = 234.52 N (downhill) The friction force F_friction = μF_normal, where F_normal is the normal force acting on the skier perpendicular to the incline. Since the skier is not moving vertically, this force equals the vertical component of gravity acting on the skier: F_normal = mg * cos(angle) = (70 kg)(9.81 m/s^2) * cos(20°) = 657.11 N Now, we can find the friction force: F_friction = (0.100)(657.11 N) = 65.71 N (uphill) The net force acting on the skier is the difference between the gravitational and frictional forces: F_net = 234.52 N - 65.71 N = 168.81 N (downhill) We can now find the acceleration of the skier using Newton's second law (F_net = ma): a = F_net/m = 168.81 N / 70 kg = 2.411 m/s^2 (downhill) Since the acceleration acts opposite to the direction of motion, we will use the following kinematic equation to find the distance up the incline: v^2 = u^2 - 2ad Where v is the final velocity (0 m/s), u is the initial velocity (4.50 m/s), a is the acceleration found (2.411 m/s^2), and d is the distance traveled. 0 = (4.50 m/s)^2 - 2(2.411 m/s^2)d Solve for d: d = (4.50 m/s)^2 / (2 * 2.411 m/s^2) = 4.218 m So, with friction considered, the skier will move 4.218 m up the incline before momentarily stopping.