A 40 kg skier comes directly down a frictionless ski slope...

A $40 \mathrm{~kg}$ skier comes directly down a frictionless ski slope that is inclined at an angle of $10^{\circ}$ with the horizontal while a strong wind blows parallel to the slope. Determine the magnitude and direction of the force of the wind on the skier if (a) the magnitude of the skier's velocity is constant, (b) the magnitude of the skier's velocity is increasing at a rate of $1.0 \mathrm{~m} / \mathrm{s}^{2} .$ and $(\mathrm{c})$ the magnitude of the skier's velocity is increasing at a rate of $2.0 \mathrm{~m} / \mathrm{s}^{2}$.

Key Concepts

Newton's Second Law of Motion

This law states that the total force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). It is fundamental in analyzing any dynamics problem as it allows one to relate forces (including both gravitational and external forces such as wind) to the resulting acceleration of an object.

Force Decomposition on an Inclined Plane

On an inclined plane, gravitational force must be split into two components: one perpendicular to the plane and one parallel to it. This technique simplifies the problem by isolating the forces that contribute to motion along the slope from those that do not, and is essential when analyzing motion on inclined surfaces.

Frictionless Surfaces

A frictionless condition in physics problems implies that there are no opposing forces due to friction. This assumption simplifies the analysis as it means only the components of gravity and any additional applied forces, such as wind, determine the acceleration along the plane.

Relationship Between Acceleration and Net Force

The net force acting on an object not only determines the direction of motion but also its rate of acceleration. Recognizing the balance of forces—especially under conditions of constant speed (where net acceleration is zero) versus positive acceleration—allows one to infer the magnitude of any additional forces that must be present to account for the observed motion.

Impact of External Forces on Motion

When an additional force (like wind) is applied, it alters the net force experienced by the object. In the context of mechanics, understanding how this extra force contributes to or opposes the gravitational component enables one to determine its magnitude and direction based on whether the motion is at constant velocity or is accelerating.

Transcript

00:01 So here we've got our skier that is sliding down the hill.

00:06 And i've over -exaggerated this angle so you can see it.

00:10 But this angle should only be 10 degrees, which isn't that great.

00:15 And we need to figure out what the magnitude of the force of the wind and its direction in three different cases.

00:29 The first case is the skier's velocity is constant, which means their net force.

00:35 Acting on it is zero, which if we draw the force diagram for this skier, they're going to have their weight force down.

00:44 And their normal force, and then a force due to the wind.

00:52 We'll call that wind force.

00:53 Now this is a skier on a frictionless incline, frictionless ski slope, so there won't be any friction.

01:00 Here so i don't really need this normal force to solve this problem, but we will need the wind force.

01:07 So the net force is zero in this case.

01:11 So for part a, the net force is zero.

01:13 The velocity is constant.

01:16 And so we say zero is equal to the sum of the forces in the parallel direction.

01:21 So that is their weight force, m .g, times the sign of that angle, which is 10 degrees, minus the force.

01:31 That the wind is applying on them.

01:33 So if our velocity is not changing, the wind force is going to be equal to, has to be equal to the parallel component of their weight force.

01:45 So the weight force is mg sine, or the wind force is mg sine of theta.

01:51 And so we can plug our numbers into that.

01:53 We're going to get a value of 68 newtons.

01:58 And this is going to be directly up the slope.

02:01 So 68 newtons parallel up the slope.

02:10 Now part b, we are told that our skier is now going to be increasing their velocity at a rate of 1 meters per second squared.

02:19 So that's their acceleration a is 1 meters per second squared.

02:24 So now our net force is no longer zero.

02:28 We're going to have ma is equal to same set of forces acting here.

02:33 M .g...

Please give Ace some feedback