College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Exercises - Page 137: 43

Answer

$F=122.5N$ uphill parallel to the inclined surface.

Work Step by Step

The component of the weight parallel to the inclined surface is responsible for providing acceleration to the block i.e. $wsin\theta=mgsin\theta$ Smallest such force applied so that the body does not accelerate is $F=mgsin\theta=25\times9.8sin30^{\circ}=122.5N$ uphill parallel to the inclined surface.
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