/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 A 60.0 -kg skier sturts from res... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 62

A 60.0 -kg skier sturts from rest at the top of a ski slope 65.0 \(\mathrm{m}\) high. (a) If frictional forces do \(-10.5 \mathrm{kJ}\) of work on her as she descends, how fast is she going at the bottom of the slope? (b) Now moving horizontally, the skier crosses a patch of soft snow, where \(\mu_{\mathrm{k}}=0.20 .\) If the patch is 82.0 \(\mathrm{m}\) wide and the average force of air resistance on the skier is 160 \(\mathrm{N}\) , how fast is she going after crossing the patch? (c) The skier hits a snowdrift and penetrates 2.5 \(\mathrm{m}\) into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Short Answer

Expert verified
(a) 24.2 m/s; (b) 12.8 m/s; (c) -1974.72 N.

Step by step solution

01

Calculate the Potential Energy

The skier starts from a height of 65.0 meters. We need to calculate the initial gravitational potential energy (PE) using the formula: \[ PE = mgh \]where \( m = 60.0 \, \text{kg} \) is the mass, \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( h = 65.0 \, \text{m} \) is the height. Substituting the values, we get:\[ PE = 60.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 65.0 \, \text{m} = 38220 \, \text{J} \]
02

Determine Kinetic Energy at the Bottom

At the bottom of the slope, potential energy converts into kinetic energy (KE), minus the work done by friction:\[ KE = PE - ext{frictional work} = 38220 \, \text{J} - 10500 \, \text{J} = 27720 \, \text{J} \]
03

Calculate Speed at the Bottom

We find the speed using the kinetic energy formula:\[ KE = \frac{1}{2}mv^2 \]Solving for \( v \):\[ 27720 \, \text{J} = \frac{1}{2} \times 60.0 \, \text{kg} \times v^2 \]\[ v^2 = \frac{27720}{30} \, \text{m}^2/\text{s}^2 \]\[ v \approx 24.2 \, \text{m/s} \]
04

Calculate Work Done by Friction and Air Resistance

While crossing the soft snow patch, calculate the total work done against friction and air resistance:- Frictional force: \( F_k = \mu_k mg = 0.20 \times 60.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 117.6 \, \text{N} \)- Total force: \( F_{total} = 117.6 \, \text{N} + 160 \, \text{N} = 277.6 \, \text{N} \)- Work done: \[ W = F_{total} \times d = 277.6 \, \text{N} \times 82.0 \, \text{m} = 22783.2 \, \text{J} \]
05

Calculate Speed After Crossing the Snow Patch

The speed after crossing can be calculated by subtracting the work done by friction and air resistance from the kinetic energy at the bottom:Initial KE: \( 27720 \, \text{J} \)Work done: \( 22783.2 \, \text{J} \)Remaining KE: \( 27720 \, \text{J} - 22783.2 \, \text{J} = 4936.8 \, \text{J} \)Use the KE formula to find the new speed \( v \):\[ 4936.8 = \frac{1}{2} \times 60.0 \, \text{kg} \times v^2 \]\[ v^2 = \frac{4936.8}{30} \]\[ v \approx 12.8 \, \text{m/s} \]
06

Calculate Average Force Exerted by the Snowdrift

To stop the skier, the work done by the snowdrift matches her final kinetic energy:- Stopping distance (d): \( 2.5 \, \text{m} \)- Work-Energy Principle: \( F_{avg} \times d = - ext{KE} \)- \( F_{avg} \times 2.5 = -4936.8 \)Solve for \( F_{avg} \):\[ F_{avg} = -\frac{4936.8}{2.5} \approx -1974.72 \, \text{N} \]The force is negative indicating that it opposes the skier's motion.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of motion. Any object that is moving has kinetic energy. It depends on two main factors: the mass of the object and its velocity. The formula to calculate kinetic energy \( KE \) is:
  • \( KE = \frac{1}{2} mv^2 \),
where \( m \) is the mass, and \( v \) is the velocity. In the exercise with the skier, potential energy is converted into kinetic energy as she comes down the slope.
This transformation is crucial to determine how fast she is going at the bottom. Since frictional forces do work on her, reducing the energy available for kinetic energy, this must be subtracted from the potential energy to find her speed.
This concept is essential for understanding how energy is conserved or transformed from one type to another as the skier moves.
Understanding kinetic energy helps us analyze situations where an object's speed changes due to forces acting on it, like friction or resistance. Knowing the speed of an object can help predict the motion of objects in various physical systems.
Potential Energy
Potential energy is the energy stored in an object due to its position or height. For instance, when an object is raised to a certain height, it possesses gravitational potential energy. This energy can be calculated with:
  • \( PE = mgh \),
where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above the ground.
In the skier exercise, she starts at the top of a slope, gaining potential energy. As she descends, this potential energy is converted into kinetic energy, determining how fast she will be when she reaches the bottom.
Understanding potential energy is critical to grasp how energy is stored and converted to kinetic energy, which helps solve real-world physics problems. When evaluating the energy exchanges, it's important to remember that potential energy can transform into other energy forms, like kinetic energy, illustrating the conservation of energy principle.
Frictional Forces
Frictional forces are forces that oppose the motion of an object. When a skier descends a slope, friction forces, including air resistance, exert against her motion, converting some of the mechanical energy into thermal energy, thus reducing her speed.
The work done by frictional forces can be calculated by multiplying the frictional force by the distance over which it acts:
  • \( W = F imes d \)
In the solution to the exercise, the impact of friction is calculated to determine how it affects the skier's kinetic energy as she moves over snow. This involves considering the coefficient of kinetic friction \( \mu_k \), and the resulting force is used to determine how much energy is lost, reducing the skier's speed.
Understanding friction is essential, as it is a common real-world force that impacts many motion scenarios. It allows us to predict how objects slow down over surfaces or through media, like snow or air, using the concepts of energy loss and conversion.

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Most popular questions from this chapter

A 2.50 -kg mass is pushed against a horizontal spring of force constant 25.0 \(\mathrm{N} / \mathrm{cm}\) on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring in any way. When the spring has been compressed enough to store 11.5 \(\mathrm{J}\) of potential energy in it, the mass is suddenly released from rest. (a) Find the greatest speed the mass reaches. When does this occur? (b) What is the greatest acceleration of the mass, and when does it occur?

CALE A certain spring is found not to obey Hooke's law; it exerts a restoring force \(F_{x}(x)=-\alpha x-\beta x^{2}\) if it is stretched or compressed, where \(\alpha=60.0 \mathrm{N} / \mathrm{m}\) and \(\beta=18.0 \mathrm{N} / \mathrm{m}^{2} .\) The mass of the spring is negligible, (a) Calculate the potential-energy function \(U(x)\) for this spring. Let \(U=0\) when \(x=0 .\) (b) An object with mass 0.900 \(\mathrm{kg}\) on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 \(\mathrm{m}\) to the right (the \(+x\) -direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 \(\mathrm{m}\) to the right of the \(x=0\) equilibrium position?

BIO Bone Fractures. The maximum energy that a bone can absorb without breaking depends on its characteristics, such as its cross-sectional area and its elasticity. For healthy human leg bones of approximately 6.0 \(\mathrm{cm}^{2}\) cross-sectional area, this energy has been experimentally measured to be about 200 \(\mathrm{J}\) . (a) From approximately what maximum height could a 60 -kg person jump and land rigidly upright on both feet without breaking his legs? (b) You are probably surprised at how small the answer to part (a) is. People obviously jump from much greater heights without breaking their legs. How can that be? What else absorbs the energy when they jump from greater heights? (Hint: How did the person in part (a) land? How do people normally land when they jump from greater heights? (c) In light of your answers to parts (a) and (b), what might be some of the reasons that older people are much more prone than younger ones to bone fractures from simple falls (such as a fall in the shower)?

CALC A proton with mass \(m\) moves in one dimension. The potential-energy function is \(U(x)=\alpha / x^{2}-\beta / x,\) where \(\alpha\) and \(\beta\) are positive constants. The proton is released from rest at \(x_{0}=\alpha / \beta .\) (a) Show that \(U(x)\) can be written as $$U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right]$$ Graph \(U(x) .\) Calculate \(U\left(x_{0}\right)\) and thereby locate the point \(x_{0}\) on the graph. (b) Calculate \(v(x)\) , the speed of the proton as a function of position. Graph \(v(x)\) and give a qualitative description of the motion. (c) For what value of \(x\) is the speed of the proton a maximum? What is the value of that maximum speed? (d) What is the force on the proton at the point in part (c)? (e) Let the proton be released instead at \(x_{1}=3 \alpha / \beta\) . Locate the point \(x_{1}\) on the graph of \(U(x) .\) Calculate \(v(x)\) and give a qualitative description of the motion. (f) For each release point \(\left(x=x_{0}\) and \(x=x_{1}\right),\) what are the maximum and minimum values of \(x\) reached during the motion?

A wooden rod of negligible mass and length 80.0 \(\mathrm{cm}\) is pivoted about a horizontal axis through its center. A white rat with mass 0.500 kg clings to one end of the stick, and a mouse with mass 0.200 kg clings to the other end. The system is released from rest with the rod horizontal. If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position?
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