Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 5 - Force and Motion-I - Problems: 1


2.9 $m/s^2$

Work Step by Step

There are two forces acting on the body so first we need to find the net force. One force (9N) is acting due east and the other is $62^{\circ}$ North of West. We must first separate the second force into its x and y components, which we can do using trigonometry. The x component, Fx is given by Fx = Fcos(62) = 3.8N West and the vertical component, Fy is given by Fy = Fsin(62) = 7.1N North. We have two forces in the x direction so we need to find the net force, which is Fnet,x = 9N (East) - 3.8N (West) = 5.2N East. We then use the equation F = ma in the x direction to find $a_{x} = F_{x}/m = $ 5.2N/3kg = 1.7 $m/s^2$ We do the same in the y direction but there is only one force so $a_{y} = F_{y}/m = $ 7.1N/3kg = 2.4 $m/s^2$ To find the magnitude of a we need to take $\sqrt (a_{x} ^2 + a_{y}^2 $ = $\sqrt (1.7^2 + 2.4^2)$ = 2.9 $m/s^2$
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