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Chapter 8: Problem 41

At a playground, a \(19-\mathrm{kg}\) child plays on a slide that drops through a height of \(2.3 \mathrm{m}\). The child starts at rest at the top of the slide. On the way down, the slide does a nonconservative work of -361 J on the child. What is the child's speed at the bottom of the slide?

Short Answer

Expert verified
The child's speed at the bottom is approximately 9.12 m/s.

Step by step solution

01

Identify Given Values

We are given the mass of the child as \( m = 19 \, \text{kg} \), the height of the slide \( h = 2.3 \, \text{m} \), and the nonconservative work done by the slide as \( W_{nc} = -361 \, \text{J} \). The initial speed of the child at the top is \( 0 \, \text{m/s} \).
02

Apply Conservation of Energy Principle

We will use the conservation of energy principle that includes nonconservative work. The total mechanical energy at the top plus any work done equals the mechanical energy at the bottom. The equation is given by:\[ \Delta KE + \Delta PE = W_{nc} \]where \( \Delta KE \) is the change in kinetic energy and \( \Delta PE \) is the change in potential energy.
03

Calculate Potential Energy Change

The change in potential energy, \( \Delta PE \), when descending the height is:\[ \Delta PE = mgh \]\[ \Delta PE = (19)(9.8)(2.3) \approx 428.34 \, \text{J} \].
04

Calculate Kinetic Energy Change

The child starts from rest, so initial kinetic energy is \( 0 \). Let \( v_f \) be the final velocity. Thus, the change in kinetic energy, \( \Delta KE \), is:\[ \Delta KE = \frac{1}{2}mv_f^2 - 0 = \frac{1}{2}mv_f^2 \]
05

Solve for Final Speed

Apply the energy conservation equation:\[ \Delta KE = W_{nc} - \Delta PE \]\[ \frac{1}{2}mv_f^2 = -361 - 428.34 \]\[ \frac{1}{2}(19)v_f^2 = -789.34 \]\[ 9.5v_f^2 = 789.34 \]\[ v_f^2 = \frac{789.34}{9.5} \approx 83.09 \]\[ v_f \approx \sqrt{83.09} \approx 9.12 \, \text{m/s} \].

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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 a form of energy that a body possesses due to its motion. In the context of our playground exercise, the child's kinetic energy is initially zero because the child begins at rest at the top of the slide. As the child slides down, energy is converted from potential to kinetic, which increases as the child speeds up.

The formula for kinetic energy (KE) is:
  • KE=\frac{1}{2}mv^2
Where:
  • \( m \) is the mass of the body.
  • \( v \) is the velocity of the body.
The kinetic energy changes as the child's velocity increases down the slide. This change in energy helps us to understand how fast the child will be going at the bottom.
Potential Energy
Potential energy is the energy stored in an object due to its position or height. At the top of the slide, the child has maximum potential energy because they are at the highest point. This energy is given by:
  • PE = mgh
Where:
  • \( m \) is the mass of the child.
  • \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).
  • \( h \) is the height of the slide.
As the child descends the slide, this potential energy is transformed into kinetic energy, due to the decrease in height and increase in speed. At the bottom, the potential energy is minimal, emphasizing the conversion from potential to kinetic energy.
Nonconservative Work
Nonconservative work refers to work done by forces that do not conserve mechanical energy, such as friction or air resistance. In this exercise, the slide performs nonconservative work on the child, with a value of -361 J. This negative work means that energy is lost or dissipated, typically as heat due to friction between the child and the slide surface.

The total mechanical energy given by the sum of kinetic and potential energy is affected by this nonconservative work. In our example, it diminishes the amount of energy available to be converted between kinetic and potential forms, meaning not all potential energy gets converted to kinetic energy during the slide. Such work often reduces the speed gained at the journey's end compared to an ideal, frictionless scenario.

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Most popular questions from this chapter

A \(17,000-\mathrm{kg}\) airplane lands with a speed of \(82 \mathrm{m} / \mathrm{s}\) on a stationary aircraft carrier deck that is \(115 \mathrm{m}\) long. Find the work done by non conservative forces in stopping the plane.

A sled slides without friction down a small, ice-covered hill. If the sled starts from rest at the top of the hill, its speed at the bottom is \(7.50 \mathrm{m} / \mathrm{s}\). (a) On a second run, the sled starts with a speed of \(1.50 \mathrm{m} / \mathrm{s}\) at the top. When it reaches the bottom of the hill, is its speed \(9.00 \mathrm{m} / \mathrm{s},\) more than \(9.00 \mathrm{m} / \mathrm{s},\) or less than \(9.00 \mathrm{m} / \mathrm{s}\) ? Explain. (b) Find the speed of the sled at the bottom of the hill after the second run.

\(A\) 15,800-kg truck is moving at \(12.0 \mathrm{m} / \mathrm{s}\) when it starts down a \(6.00^{\circ}\) incline in the Canadian Rockies. At the start of the descent the driver notices that the altitude is \(1630 \mathrm{m}\). When she reaches an altitude of \(1440 \mathrm{m},\) her speed is \(29.0 \mathrm{m} / \mathrm{s}\). Find the change in (a) the gravitational potential energy of the system and (b) the truck's kinetic energy. (c) Is the total mechanical energy of the system conserved? Explain.

Two blocks, each of mass \(m\), are connected on a frictionless horizontal table by a spring of force constant \(k\) and equilibrium length \(L\). Find the maximum and minimum separation between the two blocks in terms of their maximum speed, \(v_{\max }\), relative to the table. (The two blocks always move in opposite directions as they oscillate back and forth about a fixed position.)

BIO Compressing the Ground A running track at Harvard University uses a surface with a force constant of \(2.5 \times 10^{5} \mathrm{N} / \mathrm{m}\) This surface is compressed slightly every time a runner's foot lands on it. The force exerted by the foot, according to the Saucony shoe company, has a magnitude of \(2700 \mathrm{N}\) for a typical runner. Treating the track's surface as an ideal spring, find (a) the amount of compression caused by a foot hitting the track and (b) the energy stored briefly in the track every time a foot lands.
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