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Chapter 7: Problem 41

A \(25 \mathrm{~kg}\) child plays on a swing having support ropes that are \(2.20 \mathrm{~m}\) long. A friend pulls her back until the ropes are \(42^{\circ}\) from the vertical and releases her from rest. (a) What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing? (b) How fast will she be moving at the bottom of the swing? (c) How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

Short Answer

Expert verified
(a) 138.47 J, (b) 3.33 m/s, (c) 0 J.

Step by step solution

01

Identify the Problem Variables

Identify the given parameters in the problem. We have:- Mass of the child, \( m = 25 \text{ kg} \)- Length of the swing ropes, \( L = 2.20 \text{ m} \)- Angle of inclination, \( \theta = 42^\circ \) from the vertical.We'll be calculating potential energy, speed at the bottom of the swing, and tension work.
02

Calculate Height at Release

Use trigonometry to find the vertical height \( h \) from which the child is released. The height can be calculated as:\[ h = L(1 - \cos(\theta)) \]Substituting the known values:\[ h = 2.20 (1 - \cos(42^\circ)) \approx 2.20 (1 - 0.7431) \approx 0.5656 \text{ m} \]
03

Calculate Potential Energy

The potential energy just as the child is released is given by:\[ U = mgh \]Substitute \( m = 25 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( h = 0.5656 \text{ m} \):\[ U = 25 \times 9.8 \times 0.5656 \approx 138.47 \text{ J} \]
04

Calculate Speed at the Bottom

Using energy conservation, the potential energy at the top converts to kinetic energy at the bottom:\[ \frac{1}{2}mv^2 = mgh \]\[ v^2 = 2gh \]\[ v = \sqrt{2gh} \]Substituting the known values:\[ v = \sqrt{2 \times 9.8 \times 0.5656} \approx 3.33 \text{ m/s} \]
05

Calculate Work Done by Tension

The tension in the rope does no work as it always acts perpendicular to the direction of motion. Thus, the work done by tension is:\[ W_{\text{tension}} = 0 \text{ J} \]

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Key Concepts

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

Potential Energy
Potential energy is a type of mechanical energy that describes the energy of an object due to its position relative to other objects. In this exercise, when the child is pulled back on the swing, the height she reaches gives her gravitational potential energy. This potential energy depends on three key factors:
  • Mass of the child (\(m = 25~\text{kg}\)).
  • The height (\(h\)) from which she is released.
  • The acceleration due to gravity (\(g = 9.8~\text{m/s}^2\)).
To calculate this energy, we use the formula:\[U = mgh\]Here, height is derived from the rope length and the angle using trigonometry. Once the child is released, as she moves down, the potential energy is converted into kinetic energy.
Kinetics
Kinetics refers to the study of motion and the forces that affect it. In this problem, when the child swings down, her potential energy is converted into kinetic energy, which is the energy of motion. This conversion can be described using the equation:\[\frac{1}{2}mv^2 = mgh\]Solving for the velocity (\(v\)) at the bottom of the swing, we use the formula:\[v = \sqrt{2gh}\]This shows that the speed of the child at the bottom is dependent on the height from which she started. The concepts of kinetics help us understand and predict the motion of the child as she swings.
Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed; it can only be converted from one form to another. In this scenario, the child鈥檚 potential energy at the top of the swing is completely transformed into kinetic energy at the bottom. This is expressed by setting the initial potential energy equal to the kinetic energy when she reaches the lowest point:\[U = \frac{1}{2}mv^2\]This demonstrates that the total mechanical energy of the system remains constant throughout the swing. It informs us that no energy is lost, but rather shifted from potential to kinetic as the swing's position changes. This conservation allows us to predict the child's speed at the bottom without directly measuring it.
Trigonometry in Physics
Trigonometry is essential in physics for resolving components of vectors such as forces. In this exercise, trigonometry is used to determine the height from which the swing is released. We utilize the length of the swing and the angle from the vertical to find this height. The height (\(h\)) is calculated as:\[h = L(1 - \cos(\theta))\]where:
  • \(L\) is the length of the swing ropes.
  • \(\theta\) is the angle from the vertical.
By calculating \(1 - \cos(42^\circ)\), we convert an angular position into a linear displacement. This is crucial for all subsequent energy calculations, linking trigonometric functions to potential energy directly.

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Most popular questions from this chapter

A force of magnitude \(800.0 \mathrm{~N}\) stretches a certain spring by \(0.200 \mathrm{~m}\) from its equilibrium position. (a) What is the force constant of this spring? (b) How much elastic potential energy is stored in the spring when it is: (i) stretched \(0.300 \mathrm{~m}\) from its equilibrium position and(ii) compressed by \(0.300 \mathrm{~m}\) from its equilibrium position? (c) How much work was done in stretching the spring by the original \(0.200 \mathrm{~m} ?\)

Pendulum. A small \(0.12 \mathrm{~kg}\) metal ball is tied to a very light (essentially massless) string that is \(0.8 \mathrm{~m}\) long. The string is attached to the ceiling so as to form a pendulum. The pendulum is set into motion by releasing it from rest at an angle of \(60^{\circ}\) with the vertical. (a) What is the speed of the ball when it reaches the bottom of the arc? (b) What is the centripetal acceleration of the ball at this point? (c) What is the tension in the string at this point?

BIO An exercise program. A \(75 \mathrm{~kg}\) person is put on an exercise program by a physical therapist, the goal being to burn up 500 food calories in each daily session. Recall that human muscles are about \(20 \%\) efficient in converting the energy they use up into mechanical energy. The exercise program consists of a set of consecutive high jumps, each one \(50 \mathrm{~cm}\) into the air (which is pretty good for a human) and lasting \(2.0 \mathrm{~s},\) on the average. How many jumps should the person do per session, and how much time should be set aside for each session? Do you think that this is a physically reasonable exercise session?

Ski jump ramp. You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height \(h\) from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of \(2.0 \mathrm{~m} / \mathrm{s}\) as they reach the gate. For safety, the skiers should have a speed of no more than \(30.0 \mathrm{~m} / \mathrm{s}\) when they reach the bottom of the ramp. You determine that for an \(85.0 \mathrm{~kg}\) skier with good form, friction and air resistance will do total work of magnitude \(4000 \mathrm{~J}\) on him during his run down the slope. What is the maximum height \(h\) for which the maximum safe speed will not be exceeded?

Tendons. Tendons are strong elastic fibers that attach muscles to bones. To a reasonable approximation, they obey Hooke's law. In laboratory tests on a particular tendon, it was found that, when a \(250 \mathrm{~g}\) object was hung from it, the tendon stretched \(1.23 \mathrm{~cm}\). (a) Find the force constant of this tendon in \(\mathrm{N} / \mathrm{m}\). (b) Because of its thickness, the maximum tension this tendon can support without rupturing is \(138 \mathrm{~N}\). By how much can the tendon stretch without rupturing, and how much energy is stored in it at that point?
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