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A 4600 kg helicopter accelerates upward at 2.0 m/s^2. Calculate the lift force which exerted by...

Question:

A 4600 kg helicopter accelerates upward at 2.0 m/s{eq}^2 {/eq}. Calculate the lift force which exerted by the air on the propellers.

Gravitational Force on an Object Due to Earth

The force experienced by an object due to the gravitational field of Earth is called its weight. It is measured in Newtons. We have a minimum energy to escape from the gravitational pull of Earth. That is we must exert a force that is greater than the weight of the object.

Answer and Explanation:

Given:

{eq}m=4600\ kg\\ a=2.0\ m/s^2 {/eq}

Using the Newton's second law of motion, the net force on the helicopter is,

{eq}F_{n}=ma=4600\times1.0\ kgm/s^2=9200\ N {/eq}

The gravity pulls the helicopter downwards in the form of weight. This force is,

{eq}F_w=mg=4600\times9.8\ kgm/s^2=45080\ N {/eq}

Therefore, the actual force given by the air using propellers is,

{eq}F_{l}=F_g-F_n=45080-9200\ N=35880\ N {/eq}


Learn more about this topic:

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Gravitational Pull of the Earth: Definition & Overview

from AEPA Geography (AZ004): Practice & Study Guide

Chapter 15 / Lesson 17
60K

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