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Chapter 6: Problem 32

A pole-vaulter just clears the bar at \(5.80 \mathrm{m}\) and falls back to the ground. The change in the vaulter's potential energy during the fall is \(-3.70 \times 10^{3} \mathrm{J} .\) What is his weight?

Short Answer

Expert verified
The vaulter's weight is approximately 637.93 N.

Step by step solution

01

Understand the Relationship

Potential energy, \( U \), is given by the formula \( U = mgh \) where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height. The problem provides the change in potential energy, \( \Delta U = -3.70 \times 10^3 \text{ J} \), and height, \( h = 5.80 \text{ m} \). We are asked to find the weight, which is the force due to gravity, \( W = mg \).
02

Rearrange the Formula

Given \( \Delta U = mgh - 0 \), since the final potential energy at ground level is zero, rearrange to find \( mg = \frac{\Delta U}{h} \). This allows us to solve for the weight without knowing \( m \) directly.
03

Solve for Weight

Substitute the given values into the rearranged formula: \[ mg = \frac{-3.70 \times 10^3 \text{ J}}{5.80 \text{ m}} \approx -637.93 \text{ N} \]. Weight, being a magnitude, is positive: \( W = 637.93 \text{ N} \).
04

State the Result

The weight of the vaulter, due to gravity, is approximately \( 637.93 \text{ N} \).

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Key Concepts

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

Weight Calculation
In physics, weight is the force exerted by gravity on an object. It can be calculated using the formula \( W = mg \), where \( W \) represents weight, \( m \) is mass, and \( g \) is the acceleration due to gravity. For Earth, \( g \) is approximately \( 9.81\, \text{m/s}^2 \).
  • Weight is not the same as mass; it is a force and is measured in newtons (N).
  • While mass is a scalar quantity, weight is a vector because it has both magnitude and direction.
To find weight when only given potential energy and height, as in our exercise with the pole-vaulter, we can rearrange the potential energy formula:\( \Delta U = mgh \). Thus, solving for \( mg \) becomes: \[ mg = \frac{\Delta U}{h}\]This rearranged formula calculates weight without direct measurement of mass, emphasizing the interplay between mass, gravity, and height.
Gravitational Force
Gravitational force is a fundamental force of nature acting between masses. It pulls objects toward each other, and on Earth, it is the force that gives weight to physical objects.
  • Gravitational force on Earth is characterized by the acceleration due to gravity \( g \), which is approximately \( 9.81\, \text{m/s}^2 \).
  • The force is always directed towards the center of the Earth.
This force can be calculated by the formula \( W = mg \), indicating that weight is just the gravitational force acting on an object. In our exercise, the change in the pole-vaulter's potential energy as he falls highlights gravity's role in doing work, transforming potential energy into kinetic energy.
Energy Conversion
Energy conversion refers to the transformation of energy from one form to another. During the pole-vaulter's fall, potential energy converts to kinetic energy as he loses height.
  • Potential energy is related to an object's position or height. The higher an object is, the more potential energy it has.
  • Kinetic energy pertains to motion; when the vaulter falls, his potential energy decreases and is converted to kinetic energy, speeding up his descent.
The concept of energy conservation asserts that energy cannot be created or destroyed; it can only change forms. Thus, in an ideal scenario without air resistance, the total mechanical energy (potential + kinetic) remains constant. This principle allows us to understand how energy flows in physical systems and can predict future behavior, like predicting speed just before the vaulter hits the ground.

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Most popular questions from this chapter

In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of \(1.7 \mathrm{m} / \mathrm{s}\). However, this speed is inadequate to compensate for the kinetic friction between the puck and the ice. As a result, the puck travels only one-half the distance between the players before sliding to a halt. What minimum initial speed should the puck have been given so that it reached the teammate, assuming that the same force of kinetic friction acted on the puck everywhere between the two players?

A car, starting from rest, accelerates in the \(+x\) direction as in the figure. It has a mass of \(1.10 \times 10^{3} \mathrm{kg}\) and maintains an acceleration of \(+4.60 \mathrm{m} / \mathrm{s}^{2}\) for \(5.00 \mathrm{s}\). Assume that a single horizontal force (not shown) accelerates the vehicle. Determine the average power generated by this force.

As a sailboat sails 52 m due north, a breeze exerts a constant force \(\overrightarrow{\mathbf{F}}_{1}\) on the boat's sails. This force is directed at an angle west of due north. A force \(\overrightarrow{\mathbf{F}}_{2}\) of the same magnitude directed due north would do the same amount of work on the sailboat over a distance of just \(47 \mathrm{m} .\) What is the angle between the direction of the force \(\overrightarrow{\mathbf{F}}_{1}\) and due north?

You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a distance of \(1.2 \mathrm{m}\) in a time of \(1.5 \mathrm{s}\). The readout on the display indicates that the average power you are producing is 82 W. What is the magnitude of the force that you exert on the handle?

You are moving into an apartment and take the elevator to the 6 th floor. Suppose your weight is \(685 \mathrm{N}\) and that of your belongings is \(915 \mathrm{N}\). (a) Determine the work done by the elevator in lifting you and your belongings up to the 6 th floor \((15.2 \mathrm{m})\) at a constant velocity. (b) How much work does the elevator do on you alone (without belongings) on the downward trip, which is also made at a constant velocity?
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