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Chapter 7: Problem 54

\(\mathrm{A} 60.0 \mathrm{~kg}\) skier starts from rest at the top of a ski slope \(65.0 \mathrm{~m}\) high. (a) If friction 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
The skier's speed at the bottom of the slope is approximately \( 36.6 ms^{-1} \), her speed after crossing the snow patch is about \( 33.6 ms^{-1} \), and the average force exerted on her by the snowdrift is approximately \( 806.4 N \).

Step by step solution

01

Identify Knowns and Unknowns

The known quantities are the mass of the skier (60.0 kg), the height of the slope (65.0 m), the work done by friction on the slope (-10.5 kJ), the kinetic friction coefficient on the snow patch (0.20), the width of the snow patch (82.0 m), the force of air resistance (160 N), the penetration distance into the snowdrift (2.5 m). The unknowns are the speed at the bottom of the slope, the speed after crossing the snow patch, and the average force exerted by the snowdrift.
02

Calculating Speed at the Bottom of the Slope

By applying the work-energy theorem, \(\Delta KE = W \), where \( \Delta KE \) is the change in kinetic energy and \( W \) is the work done. The initial kinetic energy is zero (the skier starts from rest), and so the change in kinetic energy is simply the final kinetic energy. The work done on her by gravity is \( W_g = mgh = 60.0 kg * 9.81 m/s^2 * 65.0 m = 38070 J \), and the work done by friction is -10500 J, so total work done \( W = W_g + W_f = 38070 J - 10500 J = 27570 J \). Therefore, the final kinetic energy is equal to the total work, which means the final speed can be found using \( KE = \frac{1}{2} m v^2 \), solving for \( v \).
03

Calculating Speed after Crossing the Snow Patch

She loses some of her energy to the work done against kinetic friction and air resistance. The work done against friction is equal to the frictional force times the distance, \( W_{fr} = \mu_k * m * g * d = 0.20 * 60.0 kg * 9.81 m/s^2 * 82.0 m \), and work against air resistance is \( W_{air} = F_{air} * d = 160 N * 82.0 m \). The total work done is \( W_{tot} = W_{fr} + W_{air} \). The skier loses this much kinetic energy, thus reducing her speed, which can be found similarly to step 1.
04

Calculating the Average Decelerating Force of the Snowdrift

The snowdrift brings the skier to a complete stop, thus the final kinetic energy is zero. The work done by the snowdrift is equal to the kinetic energy the skier had just before hitting it, as this is the amount of energy dissipated to bring her to a stop, \( W_{snow} = \frac{1}{2} * m * v^2 \), where \( v \) is speed after crossing the snow patch. \( W_{snow} \) is also equal to the average force times the distance, \( F_{avg} * d = W_{snow} \), allowing calculation of \( F_{avg} \).

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Key Concepts

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

The Principle of Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It's a form of mechanical energy and is calculated using the equation: \[ KE = \frac{1}{2} mv^2 \] where \( m \) is the object's mass and \( v \) is its velocity. In our exercise, the skier starts with zero kinetic energy at rest and gains kinetic energy as she descends the hill due to gravitational force acting on her. This energy can be used to calculate her velocity at various points during her descent and to determine how kinetic energy transforms when subjected to different forces like friction or when penetrating a snowdrift.
Understanding Frictional Force
Frictional force is the resistance force that acts opposite to the direction of motion when two surfaces are in contact. It often does negative work on a moving object, meaning it removes energy from the system. The work done by friction can be calculated using the following formula:\[ W_{fr} = \mu_k \times m \times g \times d \] where \( \mu_k \) represents the coefficient of kinetic friction, \( m \) is the object's mass, \( g \) is the acceleration due to gravity, and \( d \) is the distance over which the force acts. Our skier experiences a loss in kinetic energy due to friction on the slope and the patch of snow, which directly impacts her velocity.
The Effects of Air Resistance
Air resistance, also known as drag, is a force that opposes the motion of an object as it moves through air. It is often modeled as a force exerted opposite to an object's velocity and can be significant at high speeds. The force of air resistance must be overcome by expending energy; and in our exercise, it plays a critical role as the skier moves horizontally across the patch of snow. The work done against air resistance is given by: \[ W_{air} = F_{air} \times d \] where \( F_{air} \) is the air resistance force and \( d \) is the distance traveled. The skier's velocity after crossing the patch corresponds to the remaining kinetic energy after the work done by both air resistance and friction has been subtracted.
Average Decelerating Force
When an object slows down, it experiences a decelerating force. The average decelerating force is a useful measure that represents the constant force that would be required to stop the object over a particular distance. In the case of our skier, the snowdrift's average decelerating force is determined by the formula:\[ F_{avg} \times d = \frac{1}{2} m v^2 \]Where \( d \) is the distance the skier penetrates into the snowdrift, and \( v \) is her velocity before impact. This calculation shows the force exerted by the snowdrift that quickly dissipates her kinetic energy, bringing her to a stop.

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Most popular questions from this chapter

DATA You are designing a pendulum for a science museum. The pendulum is made by attaching a brass sphere with mass \(m\) to the lower end of a long. light metal wire of (unknown) length \(L\). A device near the top of the wire measures the tension in the wire and transmits that information to your laptop computer. When the wire is vertical and the sphere is at rest, the sphere's center is \(0.800 \mathrm{~m}\) above the floor and the tension in the wire is \(265 \mathrm{~N}\). Keeping the wire taut, you then pull the sphere to one side (using a ladder if necessary) and gently release it. You record the height \(h\) of the center of the sphere above the floor at the point where the sphere is released and the tension \(T\) in the wire as the sphere swings through its lowest point. You collect your results: \begin{tabular}{l|lllllll} \(h(\mathbf{m})\) & 0.800 & 2.00 & 4.00 & 6.00 & 8.00 & 10.0 & 12.0 \\ \hline \(\boldsymbol{T}(\mathrm{N})\) & 265 & 274 & 298 & 313 & 330 & 348 & 371 \end{tabular} Assume that the sphere can be treated as a point mass, ignore the mass of the wire, and assume that total mechanical energy is conserved through each measurement. (a) Plot \(T\) versus \(h,\) and use this graph to calculate \(L\). (b) If the breaking strength of the wire is \(822 \mathrm{~N}\), from what maximum height \(h\) can the sphere be released if the tension in the wire is not to exceed half the breaking strength? (c) The pendulum is swinging when you leave at the end of the day. You lock the museum doors, and no one enters the building until you return the next morning. You find that the sphere is hanging at rest. Using energy considerations, how can you explain this behavior?

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CALC The potential energy of two atoms in a diatomic molecule is approximated by \(U(r)=\left(a / r^{12}\right)-\left(b / r^{6}\right),\) where \(r\) is the spacing between atoms and \(a\) and \(b\) are positive constants. (a) Find the force \(F(r)\) on one atom as a function of \(r .\) Draw two graphs: one of \(U(r)\) versus \(r\) and one of \(F(r)\) versus \(r\). (b) Find the equilibrium distance between the two atoms. Is this equilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it - that is, to separate the two atoms to an infinite distance apart? This is called the dissociation energy of the molecule. (d) For the molecule CO, the cquilibrium distance between the carbon and oxygen atoms is \(1.13 \times 10^{-10} \mathrm{~m}\) and the dissociation energy is \(1.54 \times 10^{-18} \mathrm{~J}\) per molecule. Find the values of the constants \(a\) and \(b\).

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