Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 30

A 77 -kg person is parachuting and experiencing a downward acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) shortly after opening the parachute. The mass of the parachute is \(5.2 \mathrm{~kg} .\) (a) Find the upward force exerted on the parachute by the air. (b) Calculate the downward force exerted by the person on the parachute.

Short Answer

Expert verified
Upward force exerted on the parachute by the air is 1011.06 N. The downward force exerted by the person on the parachute is 754.6 N.

Step by step solution

01

Calculate total mass and weight

First we calculate the total mass of the system (person + parachute), which is \(m = 77 kg + 5.2 kg = 82.2 kg\). The weight of the system (downward force due to gravity) can be calculated using the formula \(mg\), where \(g = 9.8 m/s^2\) is the acceleration due to gravity. So, the weight is \(F_{gravity} = mg = 82.2 kg × 9.8 m/s^2 = 805.56 N\).
02

Calculate net force

Next, calculate the net force acting on the system. This is done by multiplying the total mass of the system by the given acceleration. So, the net force is \( F_{net} = m × a = 82.2 kg × 2.5 m/s^2 = 205.5 N\).
03

Find the upward force on parachute (Part a)

The net force is defined as the difference between the upward force exerted on the parachute by the air and the downward weight of the system. By rearranging the equation for net force: \(F_{net} = F_{upward}-F_{gravity}\), we can solve for the upward force: \(F_{upward} = F_{net} + F_{gravity} = 205.5 N + 805.56 N = 1011.06 N\).
04

Find the downward force exerted by the person (Part b)

The downward force exerted by the person on the parachute is simply the weight of the person. This can be calculated using the same equation for weight we've used before, but this time only using the mass of the person: \(F_{person} = m_{person} × g = 77 kg × 9.8 m/s^2 = 754.6 N\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 car traveling at \(53 \mathrm{~km} / \mathrm{h}\) hits a bridge abutment. A passenger in the car moves forward a distance of \(65 \mathrm{~cm}\) (with respect to the road) while being brought to rest by an inflated air bag. What force (assumed constant) acts on the passenger's upper torso, which has a mass of \(39 \mathrm{~kg}\) ?

An electron is projected horizontally at a speed of \(1.2 \times\) \(10^{7} \mathrm{~m} / \mathrm{s}\) into an electric field that exerts a constant vertical force of \(4.5 \times 10^{-16} \mathrm{~N}\) on it. The mass of the electron is \(9.11 \times 10^{-31} \mathrm{~kg} .\) Determine the vertical distance the electron is deflected during the time it has moved forward \(33 \mathrm{~mm}\) horizontally.

Two blocks, with masses \(m_{1}=4.6 \mathrm{~kg}\) and \(m_{2}=3.8 \mathrm{~kg}\), are connected by a light spring on a horizontal frictionless table. At a certain instant, when \(m_{2}\) has an acceleration \(a_{2}=\) \(2.6 \mathrm{~m} / \mathrm{s}^{2},(a)\) what is the force on \(m_{2}\) and \((b)\) what is the acceleration of \(m_{1}\) ?

What are the weight in newtons and the mass in kilograms of (a) a 5.00-lb bag of sugar, (b) a 240-lb fullback, and \((c)\) a 1.80-ton car? (1 ton \(=2000\) lb. \()\)

An object is hung from a spring scale attached to the ceiling of an elevator. The scale reads \(65 \mathrm{~N}\) when the elevator is standing still. ( \(a\) ) What is the reading when the elevator is moving upward with a constant speed of \(7.6 \mathrm{~m} / \mathrm{s} ?(b)\) What is the reading of the scale when the elevator is moving upward with a speed of \(7.6 \mathrm{~m} / \mathrm{s}\) and decelerating at \(2.4 \mathrm{~m} / \mathrm{s}^{2} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Fluids

Read Explanation

Mechanics and Materials

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Quantum Physics

Read Explanation

Linear Momentum

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.

Sign-up for free