/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 A hang glider moving at speed \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A hang glider moving at speed \(9.5 \mathrm{m} / \mathrm{s}\) dives to an altitude 8.2 m lower. Ignoring drag, how fast is it then moving?

Short Answer

Expert verified
The hang glider's speed after the dive is approximately 15.8 m/s.

Step by step solution

01

Understand the Problem

A hang glider initially has a speed of \(9.5 \, \text{m/s}\) and dives to a lower altitude, losing 8.2 meters in height. We need to calculate its new speed when it reaches this lower altitude, ignoring drag.
02

Apply Conservation of Energy

According to the principle of conservation of energy, the sum of kinetic energy and potential energy at the initial position must be equal to the sum of kinetic energy and potential energy at the final position. The relevant formula is: \[ KE_1 + PE_1 = KE_2 + PE_2 \] where \( KE = \frac{1}{2}mv^2 \) and \( PE = mgh \).
03

Set Up the Energy Equation

At the initial position, the kinetic energy \( KE_1 = \frac{1}{2}m(9.5)^2 \) and the potential energy \( PE_1 = mgh_1 \). At the final position, kinetic energy \( KE_2 = \frac{1}{2}m(v_2)^2 \) and potential energy \( PE_2 = mg(h_1 - 8.2) \).
04

Simplify the Equation

The masses can be canceled out as they appear on both sides of the equation, leading to: \[ \frac{1}{2}(9.5)^2 + gh_1 = \frac{1}{2}v_2^2 + g(h_1 - 8.2) \]
05

Solve for the Final Speed

Rearrange the equation to solve for \( v_2 \): \[ \frac{1}{2}v_2^2 = \frac{1}{2}(9.5)^2 + g \times 8.2 \] Simplifying this, we use \( g = 9.81 \, \text{m/s}^2 \): \[ v_2^2 = (9.5)^2 + 2 \times 9.81 \times 8.2 \] Calculate this to find \( v_2 \).
06

Calculate the Final Speed

Calculate the expression: \[ v_2^2 = 90.25 + 160.884 \] \[ v_2^2 = 251.134 \] \[ v_2 = \sqrt{251.134} \] \[ v_2 \approx 15.8 \, \text{m/s} \]

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Key Concepts

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

Kinetic Energy in Motion
When an object is in motion, it possesses kinetic energy. This energy stems from the object's velocity and mass. The formula for kinetic energy is:
  • \( KE = \frac{1}{2}mv^2 \)
Where \( m \) is mass and \( v \) is velocity.
In the scenario of the hang glider, the kinetic energy changes as the velocity changes.
Initially, the hang glider moves with a speed of \( 9.5 \, \text{m/s} \).
At this speed, the kinetic energy can be calculated using its velocity squared (\( 9.5^2 \)).
As the glider descends and its speed increases due to gravity, its kinetic energy will increase.
The conservation of energy principle shows that the loss in potential energy converts directly into kinetic energy, increasing the glider’s speed.
In practical terms, this means the higher the speed, the more kinetic energy it possesses.
Understanding this will help in solving various physics problems involving moving objects.
Recognizing how velocity affects kinetic energy aids in analyzing energy transformations in mechanics.
Understanding Potential Energy
Potential energy is the energy an object has by virtue of its position in a force field, mainly gravitational.
The most common scenario in physics problems involves an object in the Earth’s gravitational field, where the formula is:
  • \( PE = mgh \)
Here, \( m \) is mass, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and \( h \) is height above a reference point.
In the case of the hang glider dropping 8.2 meters, potential energy decreases as it lowers in altitude.
This loss of potential energy translates entirely to extra kinetic energy, assuming no energy is lost to air resistance (drag).
So, when the glider starts at a higher altitude, it stores more potential energy compared to when it’s closer to the ground.
This principle is crucial to understand whenever working with objects in noticeable gravitational fields, particularly in problems similar to hang gliders or roller coasters.
By calculating the changes in potential energy, it’s possible to predict changes in speed or kinetic energy, maintaining a consistent approach to energy conservation.
Solving Physics Problems
Physics problems often require applying fundamental principles like the conservation of energy.
When solving such problems, especially those involving kinetic and potential energy, follow these steps:
  • Identify and write down known quantities and what needs to be calculated.
  • Use principles like energy conservation, understanding that the total mechanical energy of a system remains constant if only conservative forces act on it.
In our hang glider exercise, the glider starts with a certain amount of kinetic and potential energy.
By diving to a lower altitude, it gains more speed, showcased by solving for its final velocity.
To achieve this:
  • Set up an equation representing the initial and final total energy, \( KE_1 + PE_1 = KE_2 + PE_2 \).
  • Simplify by canceling out terms like mass if they appear on both sides.
  • Re-arrange the equation to solve for the unknown variable.
  • Substitute the known values to find the solution.
Using these structured steps, any physics problem becomes manageable.
It's all about identifying the physical principles at play and applying them systematically.

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Most popular questions from this chapter

In an adventure movie, a 62.5 -kg stunt woman falls \(8.10 \mathrm{m}\) and lands in a huge air bag. Her speed just before she hit the air bag was $10.5 \mathrm{m} / \mathrm{s} .$ (a) What is the total work done on the stunt woman during the fall? (b) How much work is done by gravity on the stunt woman? (c) How much work is done by air resistance on the stunt woman? (d) Estimate the magnitude of the average force of air resistance by assuming it is constant throughout the fall.
A gymnast of mass \(52 \mathrm{kg}\) is jumping on a trampoline. She jumps so that her feet reach a maximum height of \(2.5 \mathrm{m}\) above the trampoline and, when she lands, her feet stretch the trampoline down \(75 \mathrm{cm} .\) How far does the trampoline stretch when she stands on it at rest? [Hint: Assume the trampoline obeys Hooke's law when it is stretched.]
A block of wood is compressed 2.0 nm when inward forces of magnitude $120 \mathrm{N}$ are applied to it on two opposite sides. (a) Assuming Hooke's law holds, what is the effective spring constant of the block? (b) Assuming Hooke's law still holds, how much is the same block compressed by inward forces of magnitude \(480 \mathrm{N} ?\) (c) How much work is done by the applied forces during the compression of part (b)?
In Section \(6.2,\) Rosie lifts a trunk weighing \(220 \mathrm{N}\) up $4.0 \mathrm{m} .\( If it take her \)40 \mathrm{s}$ to lift the trunk, at what average rate does she do work?
Sean climbs a tower that is \(82.3 \mathrm{m}\) high to make a jump with a parachute. The mass of Sean plus the parachute is \(68.0 \mathrm{kg} .\) If \(U=0\) at ground level, what is the potential energy of Sean and the parachute at the top of the tower?
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