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Chapter 6: Problem 42

Sam, whose mass is \(75 \mathrm{kg}\), takes off down a 50 -m-high, \(10^{\circ}\) slope on his jet-powered skis. The skis have a thrust of \(200 \mathrm{N}\). Sam's speed at the bottom is \(40 \mathrm{m} / \mathrm{s}\). What is the coefficient of kinetic friction of his skis on snow?

Short Answer

Expert verified
The calculation involves the effects of gravity, friction, and the thrust from the skis on Sam's motion. After breaking these down into components and writing them in terms of Newton's second law, the acceleration is determined from the final velocity. The value for acceleration is substituted back into the force equation, and rearranged to solve for the coefficient of kinetic friction. This step-by-step process will yield the answer to the problem.

Step by step solution

01

Calculate the components of the weight parallel and perpendicular to the slope

We first resolve the weight into components parallel and perpendicular to the incline. The force acting down the slope due to Sam's weight can be calculated using \( F_{\text{gravity}} = mg \sin(伪) \), and the force perpendicular to the slope can be calculated using \( F_{\text{normal}} = mg \cos(伪) \), where g is the acceleration due to gravity (approximately 9.8 m/s虏). Because the angle is in degrees, make sure to convert it to radians before substituting it into the sine and cosine functions.
02

Determine the net force acting on Sam as he skis down the slope

The net force (F_net) acting on Sam as he skis down the slope is the force of gravity acting down the slope, plus the force of the skis (thrust), minus the force of friction. Frictional force can be calculated as \( F_{\text{friction}} = 渭_k F_{\text{normal}} \). This gives us the equation \( F_{\text{net}} = F_{\text{gravity}} + T - F_{\text{friction}} = ma \), where a is the acceleration. We can simplify this equation, isolate and solve for the unknown, 渭_k.
03

Calculate acceleration a from the final velocity

First, we use the kinematics equation \( v_f^2 = v_i^2 + 2as \) to find the acceleration (a). In this equation, s is the distance traveled down the slope, which is 50 m. Since Sam starts from rest, his initial velocity (v_i) is 0. By plugging in these values and rearranging, we can calculate the acceleration.
04

Substitute the values back into the force equation to solve for 碌k

After calculating the acceleration, substitute the values back into the force equation obtained in Step 2. Then, isolate 渭_k on one side and solve for it by dividing both sides by \( F_{\text{normal}} \). This will give the coefficient of kinetic friction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Friction on an Incline
Understanding kinetic friction on an incline is crucial for solving problems involving objects moving down slopes. Kinetic friction is the force that resists the motion of two surfaces sliding against each other. On an incline, this force acts opposite to the component of the object's weight that runs parallel to the slope.

When an object like Sam's skis slides down a slope, the kinetic friction depends on two factors: the coefficient of kinetic friction, denoted as \(\mu_k\), and the normal force, denoted as \(F_{\text{normal}}\). The formula for kinetic friction on an incline is \(F_{\text{friction}} = \mu_k F_{\text{normal}}\). The normal force is the component of the object's weight that is perpendicular to the slope, calculated using the cosine function relative to the angle of the incline.

In the given problem, to find the coefficient of kinetic friction, the normal force was determined by calculating Sam's weight (mass times gravity) and its component perpendicular to the slope through the angle's cosine. Then, the friction force was factored into determining the net force that affected Sam's acceleration down the slope.
Force of Gravity on a Slope
The force of gravity on a slope plays a significant role in determining the motion of objects moving along the slope. This force is a component of the object's weight which directly pulls the object down the incline. The weight of an object (\(mg\)) can be thought of as acting straight down, but when dealing with an incline, it's useful to break this force into two components: parallel and perpendicular.

The formula for calculating the force of gravity parallel to the slope is \(F_{\text{gravity}} = mg \sin(\alpha)\), where \(\alpha\) is the angle of the incline and \(g\) represents the acceleration due to gravity. In our scenario with Sam, this component of his weight contributes to his movement down the 50--m-high, \(10^\circ\) slope. It's important to convert angles to radians when working with trigonometric functions in physics.

This gravitational component, combined with other forces such as the skiing thrust and friction, ultimately determines Sam's acceleration and velocity down the slope.
Jet-Powered Skis Thrust
In the context of physics problems, the concept of thrust comes into play when there is a force propelling an object forward - such as with Sam's jet-powered skis. Thrust is a mechanical force, so it is produced by a system exerting force directly through some kind of exhaust or reaction element, like a jet or a propeller.

For Sam鈥檚 jet-powered skis, the thrust adds to the overall net force propelling him down the slope, and it's given as \(200 \mathrm{N}\). This thrust counteracts not only gravity but also the kinetic friction opposing the skis' movement. When calculating the net force, the thrust is a crucial component and is added to the force of gravity down the slope before accounting for friction. This propulsive force is a key factor in allowing Sam to reach the velocity of \(40 \mathrm{m} / \mathrm{s}\) by the time he reaches the bottom of the incline.
Acceleration Due to Gravity
The acceleration due to gravity is a constant force of attraction that the Earth exerts on objects towards its center. This acceleration is denoted by \(g\) and has a standard value of approximately \(9.8 \mathrm{m/s^2}\). In physics problems involving motion, \(g\) is often used to calculate various forces and ultimately determine the acceleration of an object.

In Sam鈥檚 case, solving for \(g\) was crucial to finding out the different forces acting on him as he accelerated down the slope. Specifically, \(g\) was used to calculate both the parallel and perpendicular components of his weight, influencing the gravitational force and the normal force, respectively. Additionally, the acceleration due to gravity provided the basis for determining Sam's final velocity using the kinematic equation for accelerated motion down the slope, enabling the calculation of the coefficient of kinetic friction. It's always important to remember that while \(g\) is approximately constant, the actual acceleration of an object like Sam on skis can be influenced by other factors, such as incline and friction.

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Most popular questions from this chapter

You've been called in to investigate a construction accident in which the cable broke while a crane was lifting a \(4500 \mathrm{kg} \mathrm{con}\) tainer. The steel cable is \(2.0 \mathrm{cm}\) in diameter and has a safety rating of \(50,000 \mathrm{N}\). The crane is designed not to exceed speeds of \(3.0 \mathrm{m} / \mathrm{s}\) or accelerations of \(1.0 \mathrm{m} / \mathrm{s}^{2},\) and your tests find that the crane is not defective. What is your conclusion? Did the crane operator recklessly lift too heavy a load? Or was the cable defective?

You've always wondered about the acceleration of the elevators in the 101 -story-tall Empire State Building. One day, while visiting New York, you take your bathroom scale into the elevator and stand on it. The scale reads 150 lb as the door closes. The reading varies between 120 lb and 170 1 b as the elevator travels. 101 floors. What conclusions can you draw?

An object moving in a liquid experiences a linear drag force: \(\vec{D}=(b v, \text { direction opposite the motion }),\) where \(b\) is a constant called the drag coefficient. For a sphere of radius \(R,\) the drag constant can be computed as \(b=6 \pi \eta R,\) where \(\eta\) is the viscosity of the liquid. a. Use what you've learned in calculus to prove that $$a_{x}=v_{x} \frac{d v_{x}}{d x}$$ b. Find an algebraic expression for \(v_{2}(x),\) the \(x\) -component of velocity as a function of distance traveled, for a spherical particle of radius \(R\) and mass \(m\) that is shot horizontally with initial speed \(v_{0}\) through a liquid of viscosity \(\eta\) c. Water at \(20^{\circ} \mathrm{C}\) has viscosity \(\eta=1.0 \times 10^{-3} \mathrm{Ns} / \mathrm{m}^{2} .\) Suppose a 1.0 -cm-diameter, \(1.0 \mathrm{g}\) marble is shot horizontally into a tank of \(20^{\circ} \mathrm{C}\) water at \(10 \mathrm{cm} / \mathrm{s}\). How far will it travel before stopping?

A rifle with a barrel length of \(60 \mathrm{cm}\) fires a \(10 \mathrm{g}\) bullet with a horizontal speed of \(400 \mathrm{m} / \mathrm{s}\). The bullet strikes a block of wood and penetrates to a depth of \(12 \mathrm{cm} .\) a. What resistive force (assumed to be constant) does the wood exert on the bullet? b. How long does it take the bullet to come to rest? c. Draw a velocity-versus-time graph for the bullet in the wood.

Sam, whose mass is 75 kg, takes off across level snow on his jet-powered skis. The skis have a thrust of \(200 \mathrm{N}\) and a coefficient of kinetic friction on snow of \(0.10 .\) Unfortunately, the skis run out of fuel after only 10 s. a. What is Sam's top speed? b. How far has Sam traveled when he finally coasts to a stop?
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