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Chapter 7: Problem 33

A 62.0 -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 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 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) 637.79 J

Step by step solution

01

Identify Known Values

First, we identify the given information:- Mass of skier, \(m = 62.0 \ \text{kg}\).- Initial velocity, \(v_i = 6.50 \ \text{m/s}\).- Distance of rough patch, \(d = 3.50 \ \text{m}\).- Coefficient of kinetic friction, \(\mu_k = 0.300\).- Height of hill, \(h = 2.50 \ \text{m}\).
02

Calculate Friction Force

The kinetic friction force \(F_k\) can be determined using the formula:\[F_k = \mu_k \cdot m \cdot g\]Where \(g = 9.81\ \text{m/s}^2\) is the acceleration due to gravity. Plugging in the values,\[F_k = 0.300 \times 62.0 \times 9.81 = 182.226 \ \text{N}\]
03

Work Done by Friction

The work done by the friction force is computed as:\[W_f = F_k \cdot d\]Using the values from earlier,\[W_f = 182.226 \times 3.50 = 637.791\ \text{J}\]
04

Calculate Skier's Speed After Rough Patch

The energy lost due to friction is equal to the change in the skier's kinetic energy:\[\Delta KE = W_f\]Thus, we find the final speed \(v_f'\) after the patch using:\[\frac{1}{2} m v_f'^2 = \frac{1}{2} m v_i^2 - W_f\]\[v_f'^2 = v_i^2 - \frac{2 W_f}{m}\]\[v_f'^2 = 6.50^2 - \frac{2 imes 637.791}{62}\]\[v_f'^2 = 42.25 - 20.57\]\[v_f' = \sqrt{21.68} \approx 4.66\ \text{m/s}\]
05

Calculate Speed at Bottom of Hill

Using conservation of energy, the potential energy at the top is converted to kinetic energy at the bottom:\[mgh = \frac{1}{2}m(v_f^2 - v_f'^2)\]\[v_f^2 = v_f'^2 + 2gh\]\[v_f^2 = 4.66^2 + 2 \times 9.81 \times 2.50\]\[v_f^2 = 21.68 + 49.05\]\[v_f = \sqrt{70.73} \approx 8.41\ \text{m/s}\]
06

Calculate Internal Energy Generated

The internal energy generated is equivalent to the work done against friction, which was previously calculated:\[\text{Internal energy} = 637.791 \ \text{J}\]

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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 an object possesses due to its motion. Imagine you're skiing down a slope; as you increase your speed, you'll notice you're gaining energy thanks to your movement. This change is described by the formula:
  • \( KE = \frac{1}{2} mv^2 \)
Here, \( m \) is the mass of the object and \( v \) is the velocity. As we saw in the exercise, the skier's initial kinetic energy was calculated based on her initial speed of 6.5 m/s.
Moving through the rough patch, her speed was affected by friction, reducing her kinetic energy to 4.66 m/s at the end of the patch. This loss in kinetic energy is due to the work done by the friction force, demonstrating how movement and external forces are interlinked.
Potential Energy
Potential energy is the stored energy of position, often related to the gravitational pull exerted by the Earth. When the skier reached the top of the hill, she possessed potential energy due to her height. This is calculated with:
  • \( PE = mgh \)
Where \( m \) is mass, \( g \) is the gravitational acceleration \( 9.81 \text{m/s}^2 \), and \( h \) is the height.
In the given problem, the skier had a height of 2.50 m, giving her additional potential energy at the top of the hill. Then, as she descended, this stored energy was converted back into kinetic energy, increasing her speed when she reached the bottom.
This demonstrates how position can store energy to be released later, playing a critical role in motion and speed.
Conservation of Energy
The principle of conservation of energy states that energy can neither be created nor destroyed; it can only be converted from one form to another. This law is fundamental in understanding how energy shifts between kinetic and potential forms, like seen with our skier.
Initially, the skier's energy was purely kinetic as she glided over the frictionless snow. When encountering the rough patch, some of her kinetic energy was converted into internal energy due to friction work, decreasing her speed.
  • The friction force removed energy from the system, illustrating energy transformation due to external forces.
Upon descending the hill, potential energy gained from climbing up was transformed back into kinetic energy, speeding up the skier. Thus, we observe how energy is conserved and transformed, balancing between the skier鈥檚 movement and her height.

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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 for 3.20 \(\mathrm{J}\) of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then drop a 1.20 -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.

A crate of mass \(M\) starts from rest at the top of a frictionless ramp inclined at an angle \(\alpha\) above the horizontal. Find its speed at the bottom of the ramp, a distance \(d\) from where it started. Do this in two ways: (a) Take the level at which the potential energy is zero to be at the bottom of the ramp with y positive upward.(b) Take the zero level for potential energy to be at the top of the ramp with \(y\) positive upward. (c) Why did the normal force not enter into your solution?

A 2.00 -kg block is pushed against a spring with negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m},\) compressing it 0.220 \(\mathrm{m}\) . When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37.0" (Fig. \(\mathrm{P7} .42\) ). (a) What is the speed of the block as it slides along the horizontal surface after having left the spring? (b) How far does the block travel up the incline before starting to slide back down?

A hydroelectric dam holds back a lake of surface area \(3.0 \times 10^{6} \mathrm{m}^{2}\) that has vertical sides below the water level. The water level in the lake is 150 \(\mathrm{m}\) above the base of the dam. When the water passes through turbines at the base of the dam, its mechanical energy is converted to electrical energy with 90\(\%\) efficiency. (a) If gravitational potential energy is taken to be zero at the base of the dam, how much energy is stored in the top meter of the water in the lake? The density of water is 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) . (b) What volume of water must pass through the dam to produce 1000 kilowatt-hours of electrical energy? What distance does the level of water in the lake fall when this much water passes through the dam?

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?
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