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A \(62.0 \mathrm{~kg}\) skier is moving at \(6.50 \mathrm{~m} / \mathrm{s}\) on a frictionless, horizontal snow-covered plateau when she encounters a rough patch \(3.50 \mathrm{~m}\) long. The coefficient of kinetic friction between this patch and her skis is 0.300 . After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill \(2.50 \mathrm{~m}\) high. (a) How fast is the skier moving when she gets to the bottom of the hill? (b) How much internal energy was generated in crossing the rough patch?

Short Answer

Expert verified
(a) 8.41 m/s; (b) 637.98 J.

Step by step solution

01

Calculate the Work Done by Friction

The work done by friction can be calculated using the formula: \[ W = f_k \times d \]where \( f_k = \mu_k \times N \), \( \mu_k \) is the coefficient of kinetic friction, and \( N \) is the normal force. On a horizontal surface, \( N = mg \). Given: \( \mu_k = 0.300 \), \( m = 62.0 \text{ kg} \), \( d = 3.50 \text{ m} \), \( g = 9.8 \text{ m/s}^2 \). The frictional force: \[ f_k = \mu_k \times mg = 0.300 \times 62 \times 9.8 = 182.28 \text{ N} \]The work done by friction over the 3.50 m patch is: \[ W = f_k \times d = 182.28 \times 3.50 = 637.98 \text{ J} \]
02

Calculate the Skier's Speed After the Rough Patch

The energy lost to friction will be equal to the work done by friction: Using conservation of energy: \[ \frac{1}{2} m v_i^2 - W = \frac{1}{2} m v_f^2 \]where \( v_i \) is the initial speed (6.50 m/s), and \( v_f \) is the final speed after the patch. Simplifying: \[ \frac{1}{2} \times 62.0 \times 6.50^2 - 637.98 = \frac{1}{2} \times 62.0 \times v_f^2 \]\[ 1309.25 - 637.98 = 31v_f^2 \]\[ 671.27 = 31 v_f^2 \]\[ v_f^2 = \frac{671.27}{31} \approx 21.65 \]\[ v_f = \sqrt{21.65} \approx 4.65 \text{ m/s} \]
03

Calculate the Speed at the Bottom of the Hill

Use conservation of mechanical energy, integrating the potential energy at the top and kinetic energy throughout the motion.Given that the hill is 2.50 m high:\[ mgh + \frac{1}{2} m v_f^2 = \frac{1}{2} m v_b^2 \]where \( v_b \) is the speed at the bottom of the hill, and previously found \( v_f = 4.65 \text{ m/s} \).Substitute the known values:\[ 62.0 \times 9.8 \times 2.50 + \frac{1}{2} \times 62 \times 4.65^2 = \frac{1}{2} \times 62 \times v_b^2 \]\[ 1523 + 670.16 = 31 v_b^2 \]\[ 2193.16 = 31 v_b^2 \]\[ v_b^2 = \frac{2193.16}{31} \approx 70.75 \]\[ v_b = \sqrt{70.75} \approx 8.41 \text{ m/s} \]
04

Calculate the Internal Energy Generated

The internal energy generated due to friction is equal to the work done by friction, which was calculated in Step 1: The internal energy generated is: \[ W = 637.98 \text{ J} \]

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Key Concepts

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

Kinetic Friction
When a skier encounters a rough patch while skiing, kinetic friction comes into play. This type of friction occurs when two surfaces are sliding past each other. It opposes the motion, meaning that it acts in the opposite direction the skier is moving.
In our scenario, the coefficient of kinetic friction, often denoted by \( \mu_k \), measures the frictional force created between the skier's skis and the patch. This coefficient depends on the materials of both surfaces. It's important to note that on a horizontal surface, the normal force \( N \) is equal to the weight of the skier, \( mg \), where \( m \) is mass and \( g \) is the acceleration due to gravity.
The frictional force \( f_k \) can be calculated as \( f_k = \mu_k \times N \). Given the coefficient \( \mu_k = 0.300 \), you can use this value to compute the force stopping the skier. For instance, in our exercise, this frictional force amounted to 182.28 N, which significantly affects the skier's speed on the rough patch.
Mechanical Energy Conservation
The principle of conservation of mechanical energy states that when only conservative forces (like gravity) do work, the total mechanical energy of a system remains constant.
This concept is observed when the skier moves from the top to the bottom of the hill after encountering the frictionless surfaces again. Initially, the skier has potential energy given by \( mgh \), where \( h \) is the height of the hill, and kinetic energy from her movement. As she moves downhill, potential energy transforms into kinetic energy, increasing her speed.
In problems like this, mechanical energy conservation simplifies calculations of speed and energy changes, as observed when calculating the skier's speed at the bottom of the hill. For example, potential energy is transformed, leading to increased kinetic energy and a final speed of 8.41 m/s.
Work-Energy Principle
The work-energy principle relates the work done by forces on an object to a change in its kinetic energy. It is a key concept for solving problems involving energy and motion.
In our exercise, this principle is used to consider how kinetic friction affects the skier's speed. Work \( W \) done by friction is used to calculate the reduction in kinetic energy, leading to a lower speed after crossing the rough patch. This is expressed as \( \frac{1}{2} mv_i^2 - W = \frac{1}{2} mv_f^2 \), where \( v_i \) and \( v_f \) are initial and final speeds, respectively.
The difference in energy, calculated to be 637.98 J, results in a kinetic energy drop, slowing the skier to 4.65 m/s. The work-energy principle here provides a clear path from initial conditions to a final outcome.
Internal Energy Generation
As the skier crosses the rough patch, internal energy is generated due to the work done by friction. This type of energy refers to the increase in energy within an object, often manifesting as heat due to the frictional interactions.
In practical terms, this means part of the skier's initial kinetic energy is converted into thermal energy, heating both the surface and the skis slightly. This energy loss is represented by the work done by friction, calculated earlier as 637.98 J.
Understanding internal energy generation helps explain why the skier experiences a drop in speed despite having no apparent loss of external energy. It demonstrates how energy conservation manifests in real-world scenarios, aiding in a deeper comprehension of the effects of friction.

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Most popular questions from this chapter

A spring of negligible mass has force constant \(k=1600 \mathrm{~N} / \mathrm{m}\). (a) How far must the spring be compressed so that \(3.20 \mathrm{~J}\) of potential energy is stored in it? (b) You place the spring vertically with one end on the floor. You then drop a \(1.20 \mathrm{~kg}\) book onto it from a height of \(0.80 \mathrm{~m}\) above the top of the spring. Find the maximum distance the spring will be compressed.

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A fisherman reels in \(12.0 \mathrm{~m}\) of line while landing a fish, using a constant forward pull of \(25.0 \mathrm{~N}\). How much work does the tension in the line do on the fish?

A bungee cord is \(30.0 \mathrm{~m}\) long and, when stretched a distance \(x\), it exerts a restoring force of magnitude \(k x\). Your father-in-law (mass \(95.0 \mathrm{~kg}\) ) stands on a platform \(45.0 \mathrm{~m}\) above the ground, and one end of the cord is tied securely to his ankle and the other end to the platform. You have promised him that when he steps off the platform he will fall a maximum distance of only \(41.0 \mathrm{~m}\) before the cord stops him. You had several bungee cords to select from, and you tested them by stretching them out, tying one end to a tree, and pulling on the other end with a force of \(380.0 \mathrm{~N}\). When you do this, what distance will the bungee cord that you should select have stretched?

A ball is thrown upward with an initial velocity of \(15 \mathrm{~m} / \mathrm{s}\) at an angle of \(60.0^{\circ}\) above the horizontal. Use energy conservation to find the ball's greatest height above the ground.

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