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Chapter 5: Problem 47

A skier starts from rest at the top of a hill that is inclined \(10.5^{\circ}\) with respect to the horizontal. The hillside is \(200 \mathrm{~m}\) long, and the coefficient of friction between snow and skis is \(0.0750\). At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest?

Short Answer

Expert verified
To provide the skier's distance on the level snow, replace the known values in the derived formula (like masses, gravity, friction, incline, etc.) into the equations in Steps 1, 2, and 3. Remember, this is a dynamic problem so masses will cancel out. Perform the computations for each step, and that will yield the answer.

Step by step solution

01

Calculate acceleration on inclined surface

First, calculate the acceleration of the skier on the inclined surface. This involves determining the gravitational force \(mg \sin(\theta)\) acting down the slope, and the frictional force \(mg \cos(\theta) \cdot \mu_k\) acting against the skier. Subtract the frictional force from the gravitational force and divide by the mass to find acceleration, \(a_{incline}\). Here, \(m\) is the mass of the skier (which we don't need since it cancels out), \(g\) is the acceleration due to gravity \(9.8 ms^{-2}\), \(\theta\) is the incline angle \(10.5^{\circ}\), and \(\mu_k\) is the coefficient of kinetic friction \(0.0750\).
02

Determine speed after traveling down the incline

Since the skier starts from rest, we can calculate the final velocity \(v_{incline}\) at the bottom of the hill using the equation of motion: \(v_{incline} = \sqrt{2a_{incline} d_{incline}}\), where \(d_{incline} = 200 \mathrm{~m}\) is the distance down the slope.
03

Determine distance traveled along horizontal snow

The skier slides along the horizontal snow until they come to rest, so now we calculate the distance on the horizontal surface \(d_{horizontal}\) using the equation of motion and knowing that final speed \(v_{horizontal} = 0\). The force of friction on the level surface is again \(mg \cos(\theta) \cdot \mu_k\), but now \(\theta = 0^{\circ}\) since the surface is horizontal, giving acceleration \(a_{horizontal}\). Therefore, \(d_{horizontal} = -\frac{{v_{incline}^{2}}}{{2 a_{horizontal}}}\). Notice that \(a_{horizontal}\) is negative since it's deceleration.

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Key Concepts

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

Kinetic Friction
Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. This force acts in the opposite direction to the movement. In the skier's case, kinetic friction is essential to understand because it influences how far the skier can travel.

When the skier descends the hill, kinetic friction acts against gravity's pull. This frictional force is calculated using the formula:
  • Friction force = Normal force x coefficient of kinetic friction \( \mu_k \)
The normal force on an inclined plane is different because it depends on the angle of inclination. Kinetic friction explains why skiers slow down both when they are going down a slope and when they glide along a flat surface.
Inclined Planes
An inclined plane is a flat surface tilted at an angle to the horizontal. In physics, inclined planes help us understand motion and forces. For the skier, the hill acts as an inclined plane.

The skier's motion down the hill is affected by two primary forces: the component of gravitational force pulling them down and the normal force pushing against the slope. The gravitational force component can be expressed as:
  • \( mg \sin(\theta) \), where \( \theta \) is the angle of the incline.
Inclines change how forces act: they spread the gravitational force into two components鈥攑arallel and perpendicular to the plane. Inclined planes are crucial in making it easier to understand why moving and stopping objects behave the way they do.
Equations of Motion
Equations of motion describe an object's movement and are used to predict how objects will move under different conditions. They are essential to solve this physics problem involving the skier.

For the skier starting from rest, key equations help determine speed and distance:
  • Final velocity (\( v \)) after traveling down: \( v = \sqrt{2 \cdot a \cdot d} \), where \( a \) is acceleration and \( d \) is distance.
  • Stopping distance on flat snow: \( d = -\frac{v^2}{2a} \).
These equations tie together various physical quantities鈥攕peed, distance, time, and acceleration鈥攄emonstrating how different forces interact with motion over time.
Acceleration
Acceleration is the rate at which an object's speed changes. It's central to understanding how the skier moves. On the hill, acceleration is due to gravity pulling the skier down; on the flat snow, it's the deceleration due to kinetic friction.

To calculate acceleration on the incline:
  • Identify forces acting: gravity and friction.
  • Calculate net force: gravitational pull minus frictional force.
  • Find acceleration using \( a = \frac{F_{net}}{m} \), where \( F_{net} \) is the net force
On the horizontal surface, gravity isn't pulling the skier downhill, but friction still acts to decelerate. Understanding acceleration helps explain why and how the skier transitions from speeding downhill to stopping on level ground.
Gravitational Force
Gravitational force is a fundamental force in physics that attracts two masses toward each other. For the skier, it pulls them down the slope and accelerates their descent.

Gravitational force on an incline can be broken into two components:
  • Parallel to the incline: pulling the skier downhill with \( mg \sin(\theta) \).
  • Perpendicular to the slope: this contributes to the normal force \( mg \cos(\theta) \).
Understanding gravitational force helps explain why the skier speeds up when going downhill. On the horizontal surface, gravitational force doesn't directly cause motion, but it maintains frictional contact with the snow. It is the pivotal force that dictates motion in inclined plane problems.

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Most popular questions from this chapter

On a frozen pond, a \(10-\mathrm{kg}\) sled is given a kick that imparts to it an initial speed of \(v_{0}=2.0 \mathrm{~m} / \mathrm{s}\). The coefficient of kinctic friction between sled and ice is \(\mu_{k}=0.10 .\) Use the work-energy theorem to find the distance the sled moves before coming to rest.

A truck travels uphill with constant velocity on a highway with a \(7.0^{\circ}\) slope. A 50 -kg package sits on the floor of the back of the truck and does not slide, due to a static frictional force. During an interval in which the truck travels \(340 \mathrm{~m}\), what is the net work done on the package? What is the work done on the package by the force of gravity, the normal force, and the friction force?

A child of mass \(m\) starts from rest and slides without friction from a height \(h\) along a curved waterslide (Fig. P5.46). She is launched from a height \(h / 5\) into the pool. (a) Is mechanical energy conserved? Why? (b) Give the gravitational potential energy associated with the child and her kinetic energy in terms of \(m g h\) at the following positions: the top of the waterslide, the launching point, and the point where she lands in the pool. (c) Determinc her initial speed \(v_{0}\) at the launch point in terms of \(g\) and \(h\). (d) Determine her maximum airborne height \(y_{\max }\) in terms of \(h, g\), and the horizontal speed at that height, \(v_{0 x}\) (e) Use the \(x\) -component of the answer to part (c) to climinate \(v_{0}\) from the answer to part (d), giving the height \(y_{\text {emax }}\) in terms of \(g, h\), and the launch angle \(\theta\). (f) Would your answers be the same if the waterslide were not frictionless? Explain.

Tarzan swings on a \(30.0\) -m-long vine initially inclined at an angle of \(37.0^{\circ}\) with the vertical. What is his speed at the bottom of the swing (a) if he starts from rest? (b) If he pushes off with a speed of \(4.00 \mathrm{~m} / \mathrm{s}\) ?

A horizontal spring attached to a wall has a force constant of \(850 \mathrm{~N} / \mathrm{m} . \mathrm{A}\) block of mass \(1.00 \mathrm{~kg}\) is attached to the spring and oscillates freely on a horizontal, frictionless surface as in Active Figure \(5.20 .\) The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? (b) What are the two points of interest? (c) Find the energy stored in the spring when the mass is stretched \(6.00 \mathrm{~cm}\) from equilibrium and again when the mass passes through cquilibrium after being released from rest. (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. (e) What is the speed at the halfway point? Why isn't it half the speed at equilibrium?
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