/*! 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 34 An 80.0 -kg skydiver jumps out o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 34

An 80.0 -kg skydiver jumps out of a balloon at an altitude of 1000 \(\mathrm{m}\) and opens the parachute at an altitude of 200 \(\mathrm{m}\) . (a) Assuming that the total retarding force on the diver is constant at 50.0 \(\mathrm{N}\) with the parachute closed and constant at 3600 \(\mathrm{N}\) with the parachute open, what is the speed of the diver when he lands on the ground? (b) Do you think the skydiver will be injured? Explain. (c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 \(\mathrm{m} / \mathrm{s} ?\) (d) How realistic is the assumption that the total retarding force is constant? Explain.

Short Answer

Expert verified
Step 1 and 2 involve calculating the skydiver's final velocity using energy principles with the retarding forces, implying that the final velocity upon landing may be high enough to cause injury. For a safe landing at 5 m/s, use a similar calculation to determine the new opening height. The constant retarding force is an oversimplification due to variable factors such as air resistance changing with velocity and altitude.

Step by step solution

01

Find the final velocity with parachute closed

Use the work-energy principle to find the velocity of the skydiver just before the parachute opens. The work done by gravity minus the work done by the retarding force equals the change in kinetic energy. Assume the skydiver starts from rest, i.e., initial kinetic energy is 0. The final kinetic energy when the parachute opens at 200 m altitude can be calculated using the formula: \(KE = \frac{1}{2} mv^2\), where m is the mass of the skydiver and v is the velocity. The work done by gravity is \(W_g = mg\triangle h\), and the work done by the retarding force is \(W_r = F_r\triangle h\), where \(F_r = 50.0\ N\). We have \(W_g - W_r = \frac{1}{2} mv^2\). Solve this equation for v to find the velocity at 200 m.
02

Calculate the velocity with the parachute open

Now calculate the final velocity with the parachute open from 200 m altitude to the ground using the same work-energy principle. The work done by gravity is now \(W_g' = mg(200\ m)\), and the work done by the retarding force with the parachute open is \(W_r' = F_r'(200\ m)\), where \(F_r' = 3600\ N\). The initial kinetic energy at this stage is from the previous step. Use the equation \(W_g' - W_r' + KE_{\text{initial}} = \frac{1}{2} mv_{\text{final}}^2\) to find the final velocity when the skydiver hits the ground.
03

Analyze the likelihood of injury

Consider the final velocity of the skydiver upon landing. A safe landing speed is typically around 5 m/s or less. If the calculated final velocity is significantly higher, the skydiver may be at risk of injury.
04

Determine the parachute opening height for a safe landing

To find the height at which the parachute should be opened to achieve a final speed of 5.00 m/s, work backwards using the desired final velocity and calculate the total work needed to be done by the retarding force and gravity to reach this final speed. Use this result to determine the altitude above 200 m at which the parachute must be opened.
05

Comment on the constant retarding force assumption

Assess whether constant retarding forces of 50.0 N and 3600 N are realistic assumptions by discussing factors like changes in air density, body position, and parachute drag variability with speed, which can cause the actual retarding forces to vary throughout the descent.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding the Work-Energy Principle in Skydiving
When a skydiver leaps from an aircraft, the work-energy principle comes into play, providing a foundational concept for analyzing the skydiver's motion. This principle states that the work done by all forces acting on an object results in a change in the object's kinetic energy.

In the context of skydiving, two main forces perform work on the diver: gravity, which does positive work and increases the skydiver's kinetic energy, and air resistance or retarding force, which does negative work and decreases kinetic energy. The change in the skydiver's kinetic energy can be expressed as the difference between the work done by gravity and the work done by the retarding force.

This understanding is crucial for calculating the skydiver's speed at various points during the descent and for ensuring the safe conclusion of the jump.
Kinetic Energy Calculation for a Falling Skydiver
Kinetic energy, the energy of motion, is a key factor in the physics of skydiving. It can be calculated using the formula: \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the skydiver and \( v \) is the velocity.

By manipulating the work-energy principle equations, we can determine the velocity of the skydiver at different altitudes, specifically before and after the parachute is deployed. For instance, at the moment the parachute opens, the skydiver's kinetic energy is offset by the work done against gravity and by the retarding force due to closed parachute. To analyze the skydiver's motion thoroughly, these energy calculations are essential, especially when discussing the safety of the landing.
The Impact of Retarding Force on Skydiving
Skydiving involves a significant retarding force, mostly attributed to air resistance, which plays a crucial role in a safe descent. A retarding force acts opposite to the velocity of the skydiver and works to reduce kinetic energy, thereby decreasing speed.

With the parachute closed, the force is relatively small, allowing the skydiver to accelerate to a higher speed. Once the parachute is open, the retarding force increases substantially, causing a strong deceleration. Understanding and accurately calculating the impact of varying retarding forces is vital for predicting the skydiver's velocity and ensuring the deployment height is appropriate for a soft landing.
Skydiving Safety Analysis: Ensuring a Soft Landing
The primary goal of skydiving safety analysis is to ensure that the skydiver reaches the ground at a non-injurious speed. In our problem, the speed at landing and the correct altitude for the parachute deployment are key factors determining safety.

A landing speed less than or equal to 5 m/s is generally considered safe. This dictates the calculations for when to deploy the parachute for a deceleration to an acceptable landing speed. The constant retarding force assumption simplifies calculations but might not represent the variability experienced in real-world conditions. Accordingly, factoring potential deviations in retarding forces due to changes in air density, body positioning, and equipment is crucial for a realistic safety analysis and to mitigate the risk of injury upon landing.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A block of mass 0.250 \(\mathrm{kg}\) is placed on top of a light vertical spring of force constant 5000 \(\mathrm{N} / \mathrm{m}\) and pushed downward so that the spring is compressed by 0.100 \(\mathrm{m}\) . After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

A child's pogo stick (Fig. P8.56) stores energy in a spring with a force constant of \(2.50 \times 10^{4} \mathrm{N} / \mathrm{m} .\) At position \(A\) \(\left(x_{\mathrm{A}}=-0.100 \mathrm{m}\right),\) the spring compression is a maximum and the child is momentarily at rest. At position \(B\) \(\left(x_{\mathrm{B}}=0\right),\) the spring is relaxed and the child is moving upward. At position \(C\) , the child is again momentarily at rest at the top of the jump. The combined mass of child and pogo stick is 25.0 \(\mathrm{kg}\) . (a) Calculate the total energy of the child-stick-Earth system if both gravitational and elastic potential energies are zero for \(x=0 .\) (b) Determine \(x_{\mathrm{C}}\) . (c) Calculate the speed of the child at \(x=0 .\) (d) Determine the value of \(x\) for which the kinetic energy of the system is a maximum. (e) Calculate the child's maximum upward speed.

A person with a remote mountain cabin plans to install her own hydroelectric plant. A nearby stream is 3.00 \(\mathrm{m}\) wide and 0.500 \(\mathrm{m}\) deep. Water flows at 1.20 \(\mathrm{m} / \mathrm{s}\) over the brink of a waterfall 5.00 \(\mathrm{m}\) high. The manufacturer promises only 25.0\(\%\) efficiency in converting the potential energy of the water-Earth system into electric energy. Find the power she can generate. (Large-scale hydroelectric plants, with a much larger drop, are more efficient.)

A toy cannon uses a spring to project a 5.30 -g soft rubber ball. The spring is originally compressed by 5.00 \(\mathrm{cm}\) and has a force constant of 8.00 \(\mathrm{N} / \mathrm{m}\) . When the cannon is fired, the ball moves 15.0 \(\mathrm{cm}\) through the horizontal barrel of the cannon, and there is a constant friction force of 0.0320 \(\mathrm{N}\) between the barrel and the ball. (a) With what speed does the projectile leave the barrel of the cannon? (b) At what point does the ball have maximum speed? (c) What is this maximum speed?

A 200 -g particle is released from rest at point \(A\) along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius \(R=30.0 \mathrm{cm}\) (Fig. P8.52). Calculate (a) the gravitational potential energy of the particle-Earth system when the particle is at point \(A\) relative to point \((\mathrm{B}),\) (b) the kinetic energy of the particle at point \((\mathrm{B}),\) (c) its speed at point \((\mathrm{B}),\) and \((\mathrm{d})\) its kinetic energy and the potential energy when the particle is at point \(C\).
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Measurements

Read Explanation

Mechanics and Materials

Read Explanation

Translational Dynamics

Read Explanation

Scientific Method Physics

Read Explanation

Kinematics Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.