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Chapter 7: Problem 18

A \(61 \mathrm{~kg}\) skier on level snow coasts \(184 \mathrm{~m}\) to a stop from a speed of \(12.0 \mathrm{~m} / \mathrm{s} .\) (a) Use the work-energy theorem to find the coefficient of kinetic friction between the skis and the snow. (b) Suppose a \(75 \mathrm{~kg}\) skier with twice the starting speed coasted the same distance before stopping. Find the coefficient of kinetic friction between that skier's skis and the snow.

Short Answer

Expert verified
(a) The coefficient of kinetic friction is 0.036. (b) The coefficient is 0.16.

Step by step solution

01

Understand the Work-Energy Theorem

According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy. The skier starts with kinetic energy due to his motion and comes to a stop, which means his final kinetic energy is zero.
02

Set up the Work-Energy Equation for Part (a)

The work done against friction is equal to the initial kinetic energy of the skier. The initial kinetic energy, KE, of the skier is given by:\[KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 61 \times (12.0)^2\]The work done by friction, W, is:\[W = -f_k \cdot d = -\mu_k \times N \times d\]Where:\(f_k\) is the kinetic friction force,\(d\) is the distance (184 m),\(N\) is the normal force, equal to \(mg\) for level snow and \(\mu_k\) is the coefficient of kinetic friction.
03

Calculate the Kinetic Energy for Part (a)

Calculate the initial kinetic energy:\[KE = \frac{1}{2} \times 61 \times (12.0)^2 = 4392 \, \text{J}\]
04

Relate Work Done to Friction for Part (a)

Since initial kinetic energy equals work done by friction,\[4392 = \mu_k \times 61 \times 9.8 \times 184\]Solve for \(\mu_k\):\[\mu_k = \frac{4392}{61 \times 9.8 \times 184}\]
05

Solve for \(\mu_k\) in Part (a)

Substitute and calculate \(\mu_k\):\[\mu_k \approx 0.036\]
06

Set up the Work-Energy Equation for Part (b)

The new skier's mass is 75 kg and the initial speed is twice before, thus 24 m/s. So, \[KE = \frac{1}{2} \times 75 \times (24.0)^2 = 21600 \, \text{J}\]
07

Relate Work Done to Friction for Part (b)

Use similar setup:\[21600 = \mu_k \times 75 \times 9.8 \times 184\]Solve for \(\mu_k\):\[\mu_k = \frac{21600}{75 \times 9.8 \times 184}\]
08

Solve for \(\mu_k\) in Part (b)

Substitute and calculate \(\mu_k\):\[\mu_k \approx 0.16\]

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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 sliding motion of two surfaces in contact. In this exercise, the skier experiences kinetic friction against the snow. This force works against the skier's motion and is responsible for eventually bringing the skier to a stop.
The kinetic friction force can be calculated using the equation:
  • The friction force \(f_k = \mu_k \times N\)
  • Where \(\mu_k\) is the coefficient of kinetic friction and \(N\) is the normal force.
  • The normal force \(N\) in this scenario is equal to the weight of the skier \(mg\) because the snow is level.
By applying this understanding, we observe that friction is what slows down the skier as they coast to a stop over the distance of 184 m. The work done by friction is equal to the change in the skier's kinetic energy from their initial speed to zero.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It can be calculated using the formula:\[KE = \frac{1}{2}mv^2\]Where \(m\) is the mass of the object and \(v\) is its velocity.
  • For the 61 kg skier, the initial velocity is 12 m/s. Plugging these values into the formula gives an initial kinetic energy of 4392 Joules.
  • This kinetic energy is entirely used up as work done against the frictional force, stopping the skier.
  • Similarly, the 75 kg skier starting with a higher speed of 24 m/s has a higher initial kinetic energy of 21600 Joules.
When analyzing problems involving moving objects, calculating kinetic energy helps to understand how much work is needed to stop or change the object's motion.
Physics Problems
Physics problems often need step-by-step guidance to effectively break down and solve. This problem utilizes the work-energy theorem which relates work done to changes in energy.

Here鈥檚 how you approach such problems:
  • Identify known information, such as mass, velocity, and distance of motion.
  • Apply the relevant physics equations 鈥� in this case, for kinetic energy and friction.
  • Understand the physical scenario like level snow, which affects forces like normal force.
The solution involves calculating energy changes 鈥� from initial kinetic energy to the energy lost due to friction 鈥� and determining the coefficient of kinetic friction \(\mu_k\). Breaking problems into smaller steps can make them comprehensive and easy to tackle.
Friction Coefficient
The coefficient of kinetic friction \(\mu_k\) is a measure of how much frictional force exists between two sliding surfaces.

To calculate it, use the formula for work done by friction: \[W = \mu_k \times N \times d\]Here:
  • For part (a), using known values for the skier (mass & distance), the initial kinetic energy, and setting it equal to the work done, you can solve \(\mu_k \approx 0.036\).
  • In part (b), a higher initial speed increases kinetic energy, and the computation shows \(\mu_k \approx 0.16\).
The coefficient provides insight into the efficiency of the sliding surfaces and how much of the skier's energy is dissipated as heat or other forms due to friction. A higher \(\mu_k\) means more energy is lost through friction.
Level Snow
Level snow implies that the ground is flat without inclines or declines affecting the skier. This simplifies calculations since the gravitational force is balanced completely by the normal force.
  • The normal force \(N\) is simply the skier's weight, \(mg\).
  • Thus, \(N = 61 \times 9.8\) for the first skier and \(N = 75 \times 9.8\) for the second skier.
  • This ensures the only horizontal force at play is that of kinetic friction.
By using level snow, it鈥檚 easier to isolate and calculate the effect of kinetic friction when applying the work-energy theorem. This scenario models situations in everyday life where skis slide over relatively flat snow surfaces.

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Most popular questions from this chapter

A \(68 \mathrm{~kg}\) skier approaches the foot of a hill with a speed of \(15 \mathrm{~m} / \mathrm{s}\). The surface of this hill slopes up at \(40.0^{\circ}\) above the horizontal and has coefficients of static and kinetic friction of 0.75 and \(0.25,\) respectively, with the skis. (a) Use energy conservation to find the maximum height above the foot of the hill that the skier will reach. (b) Will the skier remain at rest once she stops, or will she begin to slide down the hill? Prove your answer.

A boat with a horizontal tow rope pulls a water skier. She skis off to the side, so the rope makes an angle of \(15.0^{\circ}\) with the forward direction of motion. If the tension in the rope is \(180 \mathrm{~N},\) how much work does the rope do on the skier during a forward displacement of \(300.0 \mathrm{~m} ?\)

The power of the human heart. The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about \(7500 \mathrm{~L}\) of blood. Assume that the work done by the heart is equal to the work required to lift that amount of blood a height equal to that of the average American female, approximately \(1.63 \mathrm{~m} .\) The density of blood is \(1050 \mathrm{~kg} / \mathrm{m}^{3} .\) (a) How much work does the heart do in a day? (b) What is the heart's power output in watts? (c) In fact, the heart puts out more power than you found in part (b). Why? What other forms of energy does it give the blood?

A \(12.0 \mathrm{~N}\) package of whole wheat flour is suddenly placed on the pan of a scale such as you find in grocery stores. The pan is supported from below by a vertical spring of force constant \(325 \mathrm{~N} / \mathrm{m}\). If the pan has negligible weight, find the maximum distance the spring will be compressed if no energy is dissipated by friction.

You throw a \(20 \mathrm{~N}\) rock into the air from ground level and observe that, when it is \(15.0 \mathrm{~m}\) high, it is traveling upward at \(25.0 \mathrm{~m} / \mathrm{s}\). Use the work-energy theorem to find (a) the rock's speed just as it left the ground and (b) the maximum height the rock will reach.
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