/*! 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 68 A 2.0-kg purse is dropped from t... [FREE SOLUTION] | 91Ó°ÊÓ

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A 2.0-kg purse is dropped from the top of the Leaning Tower of Pisa and falls 55 m before reaching the ground with a speed of 27 m/s. What was the average force of air resistance?

Short Answer

Expert verified
The average force of air resistance is approximately 6.35 N.

Step by step solution

01

Identify the Known Values

From the problem statement, we know the following values: - Mass of the purse (m) = 2.0 kg - Height (h) = 55 m - Final velocity (v) = 27 m/s - Initial velocity (u) = 0 m/s (since it is dropped) - Gravitational acceleration (g) = 9.8 m/s²
02

Calculate the Potential Energy at the Top

Use the formula for gravitational potential energy:\[ PE = mgh \]Substitute the given values:\[ PE = 2.0 \times 9.8 \times 55 \ = 1078 \, \text{Joules} \]
03

Calculate the Kinetic Energy at the Ground

Use the formula for kinetic energy:\[ KE = \frac{1}{2}mv^2 \]Substitute the given values:\[ KE = \frac{1}{2} \times 2.0 \times 27^2 = 729 \, \text{Joules} \]
04

Determine the Work Done by Air Resistance

The work done by air resistance is the difference between the initial potential energy and the final kinetic energy:\[ W = PE - KE \]\[ W = 1078 - 729 = 349 \, \text{Joules} \]
05

Calculate the Average Force of Air Resistance

Work done by a force is also given by the formula:\[ W = F_{avg} \times d \]Where \( d \) is the distance fallen, 55 m in this case. Solve for the average force:\[ 349 = F_{avg} \times 55 \]\[ F_{avg} = \frac{349}{55} \approx 6.35 \, \text{Newtons} \]

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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 fundamental concept in physics that describes the energy an object has due to its motion. It is directly proportional to the mass of the object and the square of its velocity.
When the purse in the exercise reaches the ground, we calculate its kinetic energy using the formula:
  • \( KE = \frac{1}{2}mv^2 \)
In this equation:
  • \( m \) is the mass of the object, 2.0 kg for the purse.
  • \( v \) is the velocity, which is 27 m/s when it hits the ground.
Inserting these values:
  • \( KE = \frac{1}{2} \times 2.0 \times 27^2 = 729 \, \text{Joules} \)
This energy represents how much work would be needed to bring the purse to a stop if there were no other forces, like air resistance. Kinetic energy illustrates the energy transformation as the purse falls from rest to maximum motion.
Potential Energy
Potential energy is the energy stored in an object due to its position relative to other objects. For instances like this exercise, we consider gravitational potential energy, which depends on the height of the object above the ground.
We calculate potential energy using the formula:
  • \( PE = mgh \)
Where:
  • \( m \) is the mass of the purse, 2.0 kg.
  • \( g \) is the gravitational acceleration, approximately 9.8 m/s² on Earth.
  • \( h \) is the height, which is 55 m in this case.
Substituting these values gives:
  • \( PE = 2.0 \times 9.8 \times 55 = 1078 \, \text{Joules} \)
This potential energy represents the energy stored due to the purse’s elevated position on the tower. Upon release, this energy is transformed into kinetic energy as the purse falls. The decrease in height leads to a decrease in potential energy, ultimately reaching zero at ground level.
Gravitational Acceleration
Gravitational acceleration is a crucial factor influencing the motion of falling objects. On Earth's surface, it is approximately 9.8 meters per second squared (m/s²), indicating that an object's velocity increases by 9.8 m/s for every second it falls.
In this exercise, gravitational acceleration is central to both potential and kinetic energy calculations. Initially, it contributes to the potential energy of the purse due to its height:
  • \( PE = mgh \)
The gravitational force also influences the conversion of potential energy to kinetic energy as the object falls while no additional force acts on it, except for air resistance.
Even if the concept seems simple, gravitational acceleration is foundational to understanding motion. It affects how objects speed up when dropped and influences the time it takes for them to reach the ground. All objects, regardless of mass, will experience the same acceleration due to gravity in the absence of air resistance.

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Most popular questions from this chapter

(II) A 0.140-kg baseball 35.0 m/s traveling strikes the catcher's mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What was the average force applied by the ball on the glove?

(II) What average force is required to stop a 950-kg car in 8.0 s if the car is traveling at 95 km/h?

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(I) A force of 35.0 N is required to start a 6.0-kg box moving across a horizontal concrete floor. (\(a\)) What is the coefficient of static friction between the box and the floor? (\(b\)) If the 35.0-N force continues, the box accelerates at 0.60 m/s\(^2\) What is the coefficient of kinetic friction?

Piles of snow on slippery roofs can become dangerous projectiles as they melt. Consider a chunk of snow at the ridge of a roof with a slope of 34\(^\circ\). (\(a\)) What is the minimum value of the coefficient of static friction that will keep the snow from sliding down? (\(b\)) As the snow begins to melt, the coefficient of static friction decreases and the snow finally slips. Assuming that the distance from the chunk to the edge of the roof is 4.0 m and the coefficient of kinetic friction is 0.10, calculate the speed of the snow chunk when it slides off the roof. (\(c\)) If the roof edge is 10.0 m above ground, estimate the speed of the snow when it hits the ground.

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