A 68-kg water skier is being accelerated by a ski boat on a fl...

Question

A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic friction between the skier's skis and the water surface is μₖ = 0.25 (Fig. 5–59). What is the skier's horizontal acceleration if the rope pulling the skier exerts a force of F_T = 240N on the skier at an upward angle θ = 12°?

A water skier being pulled by a boat on a lake, with force and friction details illustrated.

Accepted Answer

Hello, let's go through this practice problem. An interior decorator is rearranging furniture and needs to move a 75 kg bookcase across a hardwood floor. The coefficient of kinetic friction between the bookcase and the hardwood floor is M sub K equals 0.3. The decorator applies a force of 250 new ones by dragging to slide the bookcase, calculate the bookcase's acceleration given that the force is applied at an angle theta of five degrees to the horizontal option. A 0.3 m per second squared, B 0.5 m per second squared C 0.9 m per second squared and D 1 m per second squared. So this is a problem that's asking about the acceleration of an object. And we're given some information related to its friction, its mass, uh the angle of the force being applied to it. So this is going to be a problem where we want to analyze forces. So the first thing we're going to do is draw a free body diagram of showing the forces acting on the box. So acting downwards on the cases center of mass is the weight of the bookcase. And recall that force due to weight is equal to the mass of the object multiplied by the acceleration due to gravity acting upwards against the bookcase is the normal force from the floor shall label as N the decorator is giving the bookcase an applied force at an angle. So we'll call the applied force F sub A and this force is acting at an angle theta with the horizontal. And we'll call that horizontal component F sub A X, the X component of the applied force. A label that component with a different color. Because the Y component of that applied force, we'll call F sub A Y. And lastly opposing the motion of the box is the frictional force to the left is the force from the friction F sub K. So now that we've drawn a diagram showing all the forces acting on the crate. Now let's use our force equations to write the sum of the forces along each axis. Let's start with the horizontal axis. So the net force in the X direction. Now, according to Newton's second law, the sum of the forces is equal to the object's mass multiplied by its acceleration. So in the X direction, pointing towards the right is the X component of the applied force. So F sub A X and opposing it in the opposite direction. So I'll write it as negative is the frictional force F sub K. And according to Newton's second law, the sum of these forces is equal to the objects or the bookcases mass multiplied by its acceleration. And the whole purpose of this problem is to solve for the acceleration of the case. So let's algebraically rewrite this equation to solve it. For acceleration, we simply divide both sides of the equation by mass. So we can see that the acceleration is equal to the applied force in the horizontal direction. F sub A X minus F sub K divided by the mass M. And we're going to expand these terms out a little bit more into more variables that we actually have. Recall that when we have a force at an angle like this an angle from the horizontal, then this can be rewritten as the hypotenuse or in this case, F sub a multiplied by the cosine of the angle theta. So that's how we can rewrite F sub ax. And the frictional force can be rewritten. Recall that frictional force is equal to the coefficient of friction, new sub K multiplied by the magnitude of the normal force. This is all divided by M. Now let's move on to look at the sum of the forces along the vertical axis. So F subnet, why? So looking at our force diagram, we can see that acting upwards on the crate is the normal force plus the Y component of the applied force and then opposing it pointing downwards is the weight from the box or MG. And since the box is not accelerating or moving at all in the vertical direction, it's just always in contact with the floor. That means that there is no acceleration in the vertical direction, which means that the net force as per new in second law is zero. And because the normal force is the only variable in our acceleration equation that we do not have a value for, we should rewrite our vertical component equation to solve it for the normal force. And then plug that into our acceleration equation that we found from the horizontal forces. So again, we're gonna do some pretty basic algebra here to solve for N by simply adding MG to both sides of the equation and subtracting F sub yay from both sides of the equation. So the normal force is equal to MG minus F sub A Y which can be rewritten using trigonometry as the applied force F sub a multiplied by the sine of theta. So now let's return to our acceleration equation. Acceleration is equal to the applied force multiplied by the cosine of theta minus mu sub K multiplied by the normal force only. Now we're gonna plug in the formula for the normal force that we just found. So MG minus F sub A multiplied by the sine of theta all divided by M. Now we can simply plug in all the values that are given to us in the problem. The applied force F sub A is equal to 250 new ones multiplied by the cosine of the angle theta which is five degrees minus the coefficient of friction, which is given as 0.3 multiplied by MG. So that's mass multiplied by gravity. The mass is 75 g for the bookcase multiplied by 9.81 m per second squared for G minus the applied force of 250 newtons multiplied by the sine of the angle theta of five degrees. And all of this is divided again by the mass of the bookcase which is 75 kg. Put all this into a calculator and the acceleration turns out to be approximately 0.5 m per second squared. So that is our answer to this problem which corresponds to option B. So B is our answer and that is it for this problem. I hope this video helped you out and please consider checking out our other tutoring videos which will give you more experience with these types of problems. That's all for now. And I hope you have a lovely day. Bye bye.

Key Concepts

Newton's Second Law of Motion

Friction and Coefficient of Kinetic Friction

Components of Forces

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