Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A skier starts with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\) and skis straight down a slope with an angle of \(15.0^{\circ}\) relative to the horizontal. The coefficient of kinetic friction between her skis and the snow is \(0.100 .\) What is her speed after 10.0 s?

Short Answer

Expert verified
Answer: To find the skier's speed after 10.0 s, we will calculate the forces acting on her (gravitational force parallel to the slope and kinetic friction force), determine her acceleration due to these forces, and then use kinematic equations to find her speed. The final speed can be found using the equation \(v = 2.0 \mathrm{~m} / \mathrm{s} + (g\sin{\theta} - \mu g\cos{\theta})(10.0 \mathrm{s})\) with the given values for initial speed, slope angle, and coefficient of kinetic friction.

Step by step solution

01

Calculate gravitational force acting on the skier

First, we need to determine the gravitational force acting on the skier along the slope. The gravitational force can be broken into components that are parallel and perpendicular to the slope. The parallel component will be responsible for accelerating the skier down the slope. To find the gravitational force parallel to the slope, we will use the formula: \(F_{g\parallel} = mg\sin{\theta}\) Here, \(m\) is the skier's mass, \(g \approx 9.8 \mathrm{~m} / \mathrm{s^2}\) is the gravitational acceleration, and \(\theta = 15.0^{\circ}\) is the angle of the slope.
02

Calculate kinetic friction force acting on the skier

The kinetic friction force acting on the skier will oppose her motion down the slope. To calculate this force, we will use the formula: \(F_{f} = \mu F_{N}\) Here, \(\mu = 0.100\) is the coefficient of kinetic friction, and \(F_{N}\) is the normal force acting on the skier perpendicular to the slope. Since there is no vertical acceleration, this normal force is equal to the gravitational force component perpendicular to the slope: \(F_{N} = mg\cos{\theta}\) Thus, we get: \(F_{f} = \mu mg\cos{\theta}\)
03

Calculate net force and acceleration of the skier

Now, we calculate the net force acting on the skier along the slope. The net force is the difference between the gravitational force parallel to the slope and the kinetic friction force: \(F_{net} = F_{g\parallel} - F_{f}\) Using the equations we derived in steps 1 and 2: \(F_{net} = mg\sin{\theta} - \mu mg\cos{\theta}\) Since \(F_{net} = ma\), where \(a\) is the skier's acceleration, we can find the acceleration: \(a = g\sin{\theta} - \mu g\cos{\theta}\)
04

Use kinematic equations to find the skier's speed after 10.0 s

We are given the initial speed of the skier, \(v_{0} = 2.0 \mathrm{~m} / \mathrm{s}\), and we want to find her speed after \(t = 10.0 \mathrm{s}\). We can use the kinematic equation: \(v = v_{0} + at\) Substituting the values, we have: \(v = 2.0 \mathrm{~m} / \mathrm{s} + (g\sin{\theta} - \mu g\cos{\theta})(10.0 \mathrm{s})\) Now, we can plug in the known values of \(g\), \(\mu\), and \(\theta\) to find \(v\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Gravitational Force Components
Gravitational force, also known as weight, plays a crucial role in physics, especially in kinematics problems involving motions on inclines. To fully comprehend its effect on a body sliding down a slope, like our skier, it's essential to understand its components. Gravity acts directly downward, but when an object is on an inclined surface, this force can be split into two parts: one perpendicular to the plane, and one parallel to it.

The parallel component, denoted as \(F_{g\parallel}\), is responsible for pulling the object down the incline, while the perpendicular component, \(F_{g\perp}\), interacts with the surface to produce a normal force. The normal force is significant because it affects the frictional force that opposes the downhill motion. These components are found using trigonometric functions of the incline angle, with \(F_{g\parallel} = mg\sin{\theta}\) and the normal force as \(F_{g\perp} = mg\cos{\theta}\), where \(m\) is mass, \(g\) is the acceleration due to gravity, and \(\theta\) is the angle of the incline.
Kinetic Friction
As objects move across surfaces, they experience friction - a force that resists motion. Kinetic friction, specifically, occurs when an object is sliding over a surface. In our skier's case, the kinetic friction is what opposes her motion down the snowy slope.

It is calculated using the formula \(F_{f} = \mu F_{N}\), where \(\mu\) is the coefficient of kinetic friction and \(F_{N}\) is the normal force - the force exerted by the surface perpendicular to the object. For our skier, kinetic friction can be found by multiplying the coefficient of kinetic friction with the normal force, which as we established, is the gravitational force component perpendicular to the slope \(F_{f} = \mu mg\cos{\theta}\). The important thing to remember is that kinetic friction always acts in the direction opposite to the object's motion and its magnitude is usually less than the maximum static friction.
Kinematics Equations
To predict and analyze the motion of objects, kinematics equations are employed. These equations describe the relationships between displacement, velocity, acceleration, and time without considering the forces that cause such motion. A popular equation, especially in problems involving constant acceleration like our skier's descent, is \(v = v_{0} + at\).

In this equation, \(v\) is the final velocity, \(v_{0}\) is the initial velocity, \(a\) is acceleration, and \(t\) is time. In our problem, we used this equation along with the skier's initial speed and the acceleration (caused by the net force) to determine her speed after a certain time. Kinematics equations are powerful tools in physics, allowing us to solve for various unknowns in a given motion scenario.
Coefficient of Kinetic Friction
The coefficient of kinetic friction, symbolized by \(\mu\), is a dimensionless number which characterizes the friction between two surfaces in motion. It has no units and is specific to the pair of surfaces in contact; for instance, the skier's skis on snow. The value of \(\mu\) helps us determine the frictional force by multiplying it with the normal force exerted by a surface.

In our exercise, the coefficient of kinetic friction plays a vital role in calculating the skier's net force and subsequent acceleration. Having a lower coefficient means less resistance and faster acceleration, while a higher coefficient means the opposite. Understanding this coefficient is key to predicting how much an object will slow down or speed up due to friction when moving on a surface.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two blocks of masses \(m_{1}\) and \(m_{2}\) are suspended by a massless string over a frictionless pulley with negligible mass, as in an Atwood machine. The blocks are held motionless and then released. If \(m_{1}=3.50 \mathrm{~kg}\). what value does \(m_{2}\) have to have in order for the system to experience an acceleration \(a=0.400 g\) ? (Hint: There are two solutions to this problem.

Two blocks are stacked on a frictionless table, and a horizontal force \(F\) is applied to the top block (block 1). Their masses are \(m_{1}=2.50 \mathrm{~kg}\) and \(m_{2}=3.75 \mathrm{~kg}\). The coefficients of static and kinetic friction between the blocks are 0.456 and 0.380 , respectively a) What is the maximum applied force \(F\) for which \(m_{1}\) will not slide off \(m_{2} ?\) b) What are the accelerations of \(m_{1}\) and \(m_{2}\) when \(F=24.5 \mathrm{~N}\) is applied to \(m_{1}\) ?

Two blocks \(\left(m_{1}=1.23 \mathrm{~kg}\right.\) and \(m_{2}=2.46 \mathrm{~kg}\) ) are glued together and are moving downward on an inclined plane having an angle of \(40.0^{\circ}\) with respect to the horizontal. Both blocks are lying flat on the surface of the inclined plane. The coefficients of kinetic friction are 0.23 for \(m_{1}\) and 0.35 for \(m_{2}\). What is the acceleration of the blocks?

An elevator cabin has a mass of \(358.1 \mathrm{~kg},\) and the combined mass of the people inside the cabin is \(169.2 \mathrm{~kg} .\) The cabin is pulled upward by a cable, with a constant acceleration of \(4.11 \mathrm{~m} / \mathrm{s}^{2}\). What is the tension in the cable?

A \(423.5-\mathrm{N}\) force accelerates a go-cart and its driver from \(10.4 \mathrm{~m} / \mathrm{s}\) to \(17.9 \mathrm{~m} / \mathrm{s}\) in \(5.00 \mathrm{~s}\). What is the mass of the go-cart plus driver?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free