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Chapter 8: Problem 54

A child whose weight is \(267 \mathrm{~N}\) slides down a \(6.5 \mathrm{~m}\) playground slide that makes an angle of \(20^{\circ}\) with the horizontal. The coefficient of kinetic friction between slide and child is \(0.10\). (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of \(0.457 \mathrm{~m} / \mathrm{s}\), what is her speed at the bottom?

Short Answer

Expert verified
(a) Thermal energy transferred is the work done by friction (b) Find final speed using kinetic energy conservation.

Step by step solution

01

Understanding the Scenario

The child's weight is given as the force due to gravity acting on her, which is 267 N. She's sliding down a playground slide with a distance of 6.5 m at an angle of 20掳. The slide has a coefficient of kinetic friction of 0.10. We need to calculate the thermal energy transferred and her final speed if she starts with an initial speed.
02

Calculate the Normal Force

The normal force is the component of the child's weight perpendicular to the slide. This can be calculated as: \[ N = W \cdot \cos(\theta) = 267 \cdot \cos(20^{\circ}) \]
03

Determine the Frictional Force

The frictional force can be calculated using the formula: \[ f = \mu_k \cdot N = 0.10 \cdot (267 \cdot \cos(20^{\circ})) \]
04

Compute the Work Done by Friction

The work done by friction, which equals the thermal energy, is the product of the frictional force and the distance along the slide: \[ W_f = f \cdot d = (0.10 \cdot (267 \cdot \cos(20^{\circ}))) \cdot 6.5 \]
05

Initial Kinetic and Potential Energy

At the top of the slide, her initial potential energy: \[ PE_i = mgh = 267 \times 6.5 \times \sin(20^{\circ}) \] Her initial kinetic energy: \[ KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2} \times 267 \times \left(0.457^2\right) \]
06

Calculate Total Initial Energy

The total energy at the top is the sum of initial potential and kinetic energy: \[ E_{i} = PE_i + KE_i \]
07

Energy at the Bottom of the Slide

At the bottom of the slide, all her potential energy has been converted into kinetic energy and some energy is converted into thermal energy due to friction. So the kinetic energy at the bottom: \[ KE_f = E_{i} - W_f \]
08

Final Speed Calculation

Using the final kinetic energy, calculate her speed at the bottom: \[ \frac{1}{2}mv_f^2 = KE_f \]Solving for \(v_f\):\[ v_f = \sqrt{\frac{2 \times KE_f}{m}} \]
09

Conclusion

By following the calculations from these steps, we've determined the amount of thermal energy transferred and the child's speed at the bottom of the slide.

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Key Concepts

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

Kinetic Friction
Kinetic friction plays a crucial role in situations where two surfaces move relative to each other. In the case of the child sliding down the playground slide, kinetic friction opposes the motion of the child. This force arises due to the contact between the child and the slide's surface.
  • The coefficient of kinetic friction is given as 0.10 in this exercise.
  • The frictional force is calculated as the product of this coefficient and the normal force acting on the child.
The normal force is the force exerted by a surface to support the weight of an object resting on it, which in this scenario is a component of the child's weight perpendicular to the slide.Calculating the frictional force involves using the formula: \[ f = \, \mu_k \, \times \, N \]where:
  • \( \mu_k \) is the coefficient of kinetic friction, and
  • \( N \) is the normal force.
This frictional force results in energy being transferred from mechanical energy to thermal energy as the child slides down.
Energy Transformation
Energy transformation is a concept that explains how energy changes from one form to another as different forces act on a body. In this exercise, the child's potential and kinetic energy at the top of the slide are transformed as she reaches the bottom.
This involves:
  • The gravitational force doing work on the child, increasing her kinetic energy.
  • Kinetic friction converting some of her mechanical energy into thermal energy due to sliding.
Understanding these transformations helps in calculating the child's speed and the thermal energy at the bottom. By accounting for the work done by friction, we can find out how much total mechanical energy is converted into other forms like heat.
Kinetic and Potential Energy
Kinetic energy and potential energy are the two primary forms of mechanical energy in this situation. - **Potential Energy**: It's the energy stored due to an object's position. At the top of the slide, the child has gravitational potential energy given by: \[ PE = mgh \] - Where \( m \) is the mass of the child, \( g \) is the acceleration due to gravity, and \( h \) is the height of the slide.
- **Kinetic Energy**: It's the energy due to motion. At the beginning, the child also has initial kinetic energy from her initial speed calculated by: \[ KE = \frac{1}{2}mv^2 \] - As the child slides down, potential energy decreases while her kinetic energy increases, except for energy lost as heat due to friction.Working out these energies enables us to understand the energy distribution, initially and at any given point of her motion on the slide.
Gravitational Force
Gravitational force is a fundamental concept influencing the motion of objects. It鈥檚 the force that attracts two bodies towards each other. In this exercise, the gravitational force is what pulls the child down the slide.
Every object with mass exerts a gravitational pull, and here, it's quantified as the child's weight (267 N), which is the gravitational force acting on her throughout her descent.
  • This force is crucial since it affects both her potential energy at the start and her acceleration while moving down the slide.
  • It's also perpendicular to the normal force, which is essential in determining the corresponding frictional force.
Gravitational force ensures that potential energy is gradually transformed into kinetic energy as she moves down, overcoming friction to some extent.

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Most popular questions from this chapter

Tarzan, who weighs \(688 \mathrm{~N}\), swings from a cliff at the end of a vine \(18 \mathrm{~m}\) long (Fig. 8-28). From the top of the cliff to the bottom of the swing, he descends by \(3.2 \mathrm{~m}\). The vine will break if the force on it exceeds \(950 \mathrm{~N}\) (a) Does the vine break? (b) If it does not, what is the greatest force on it during the swing? If it does, at what angle with the vertical does it break?

You drop a \(2.00 \mathrm{~kg}\) book to a friend who stands on the ground at distance \(D=10.0 \mathrm{~m}\) below. Your friend's outstretchedhands are at distance \(d=1.50 \mathrm{~m}\) above the ground (Fig. 8-19). (a) How much work \(W_{g}\) does the gravitational force do on the book as it drops to her hands? (b) What is the change \(\Delta U\) in the gravitational potential energy of the book-Earth system during the drop? If the gravitational potential energy \(U\) of that system is taken to be zero at ground level, what is \(U\) (c) when the book is released, and (d) when it reaches her hands? Now take \(U\) to be \(100 \mathrm{~J}\) at ground level and again find (e) \(W_{g}\), (f) \(\Delta U\), (g) \(U\) at the release point, and (h) \(U\) at her hands.

You push a \(2.0 \mathrm{~kg}\) block against a horizontal spring, compressing the spring by \(12 \mathrm{~cm}\). Then you release the block, and the spring sends it sliding across a tabletop. It stops \(75 \mathrm{~cm}\) from where you released it. The spring constant is \(170 \mathrm{~N} / \mathrm{m}\). What is the block-table coefficient of kinetic friction?

Adam stretches a spring by some length. John stretches the same spring later by three times the length stretched by Adam. Find the ratio of the stored energy in the first stretch to that in the second stretch.

A cookie jar is moving up a \(40^{\circ}\) incline. At a point \(45 \mathrm{~cm}\) from the bottom of the incline (measured along the incline), the jar has a speed of \(1.4 \mathrm{~m} / \mathrm{s}\). The coefficient of kinetic friction between jar and incline is \(0.15\). (a) How much farther up the incline will the jar move? (b) How fast will it be going when it has slid back to the bottom of the incline? (c) Do the answers to parts (a) and (b) increase, decrease, or remain the same if we decrease the coefficient of kinetic friction (but do not change the given speed or location)?
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