/*! 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 9 As an Acapulco cliff diver drops... [FREE SOLUTION] | 91Ó°ÊÓ

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As an Acapulco cliff diver drops to the water from a height of \(46 \mathrm{m},\) his gravitational potential energy decreases by \(25,000 \mathrm{J}\) What is the diver's weight in new tons?

Short Answer

Expert verified
The diver's weight is approximately 543.26 newtons.

Step by step solution

01

Understand the relationship between potential energy and weight

Potential energy (PE) can be calculated using the formula \( PE = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and \( h \) is the height. The change in potential energy given is \( 25,000 \, \text{J} \), and the height is \( 46 \, \text{m} \).
02

Rearrange the formula to find mass

Rearrange the potential energy formula to solve for mass: \( m = \frac{PE}{gh} \). Substitute the known values: \( m = \frac{25,000}{9.81 \times 46} \).
03

Calculate the mass of the diver

Perform the calculation: \( m = \frac{25,000}{9.81 \times 46} \approx \frac{25,000}{451.26} \approx 55.38 \, \text{kg} \). This is the mass of the diver in kilograms.
04

Convert mass to weight in newtons

Weight can be calculated using \( F = mg \), where \( F \) is the weight in newtons. Using the mass found, \( F = 55.38 \times 9.81 \approx 543.26 \, \text{N} \). This is the weight of the diver in newtons.

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Key Concepts

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

Weight Calculation
Weight calculation is a central concept in physics, particularly when dealing with gravitational forces. Weight is essentially the force exerted by gravity on an object and is calculated using the simple formula:\[F = mg\]Here, \( F \) represents the weight of an object, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity. In most physics problems, the value of \( g \) is assumed to be approximately \( 9.81 \, \text{m/s}^2 \).
  • To find the weight, multiply the mass of the object by the gravitational acceleration.
  • The result is in newtons, which is the standard unit of force.
Weight is different from mass because while mass is a measure of how much matter an object contains, weight is a measure of how strongly gravity pulls on that matter. When we calculate weight, we are figuring out this force in newtons. This distinction is vital in physics as it helps understand the effects of gravitational forces on different bodies. When solving problems, always ensure the mass is correctly converted into the right units for accurate calculations.
Physics Formulas
The usage of physics formulas allows us to solve various real-world problems by applying mathematical relationships that describe physical phenomena. One crucial formula in physics is for gravitational potential energy (GPE):\[PE = mgh\]In this formula:
  • \( PE \) represents the potential energy in joules.
  • \( m \) is the mass in kilograms.
  • \( g \) is the acceleration due to gravity, commonly \, \( 9.81 \, \text{m/s}^2 \).
  • \( h \) is the height in meters.
This formula shows the potential energy related to the position of an object relative to a gravitational field. As in our exercise, knowing the change in gravitational potential energy allows us to rearrange the formula to solve for unknown variables, such as mass.
Such rearrangements highlight the formula’s flexibility in addressing different physics problems, involving gravitational forces, and energy transformations. Understanding these relationships is fundamental in physics as they provide a base on which more complex concepts are built. Formulas like this one, when correctly used, can unravel the characteristics of physical systems.
Energy Conversion
Energy conversion is an important and widespread concept in physics, showing how energy changes forms while maintaining its essence. In the context of gravitational potential energy problems, such as in our exercise, energy conversion is about translating stored potential energy into kinetic energy or other forms.
  • Potential energy is stored due to an object’s position or state, such as a diver about to jump from a cliff.
  • As the diver jumps, potential energy converts to kinetic energy: energy of motion.
  • Understanding how energy is converted helps explain natural occurrences, such as falling objects gaining speed.
In our example, as the cliff diver descends, the gravitational potential energy decreases by \( 25,000 \, \text{J} \). This energy doesn’t just disappear; it transforms primarily into kinetic energy, driving the diver’s increasing velocity towards the water.
Energy conversion principles emphasize the conservation of energy, a pivotal law in physics asserting that energy cannot be created or destroyed, only transformed. Recognizing these transformations helps predict subsequent states of objects and systems driven by dynamic physical processes.

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Most popular questions from this chapter

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.)

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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.

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 2.9-kg block slides with a speed of 1.6 \(\mathrm{m} / \mathrm{s}\) on a frictionless horizontal surface until it encounters a spring. (a) If the block compresses the spring \(4.8 \mathrm{cm}\) before coming to rest, what is the force constant of the spring? (b) What initial speed should the block have to compress the spring by \(1.2 \mathrm{cm} ?\)

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