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As a sailboat sails 52 m due north, a breeze exerts a constant force $\overrightarrow{\mathbf{F}}_{1}$ on the boat's sails. This force is directed at an angle west of due north. A force $\overrightarrow{\mathbf{F}}_{2}$ of the same magnitude directed due north would do the same amount of work on the sailboat over a distance of just 47 $\mathrm{m}$ . What is the angle between the direction of the force $\overrightarrow{\mathbf{F}}_{1}$ and due north?
$25^{\circ}$
Physics 101 Mechanics
Chapter 6
Work and Energy
Work
Kinetic Energy
Michael B.
April 7, 2021
.As a sailboat sails 52m due north, a breeze exerts a constant force of F1 on the boat's sails. This force is directed at an angle west of due north. A force F2 of the same magnitude directed due north would do the same amount of work on the sailboat over
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So let's make a free body diagram for the first scenario. So we have a force F one which acts at some angle data to do north. And the displacement we have is in the new north direction, which is 52 meters. So we can write down that the work done in this case is equal to F one s co sign off data, which is F one. We do not know the magnitude times 52 times course, sign off there. So that's the equation that we're gonna use in a second. Now let's come to the second scenario in the seconds and there you may have another force F to, which is also due north, and the displacement in this case is 47 meters. And the work done, we would say, is equal to F two times 47. We're told that the same amount of work is done in both cases, so we can actually say that 52 F one co sign off data is equal to 47 f two. But then we're also told that the magnitude off F one is equal to the magnitude of F to so we can actually replace F one by F too. So we can say 47 50 to 1. Co signed their eyes equal to 47 f one and we can divide both sides by F one and then we can say course. Xanterra is equal to 47 over 52 which would give me theta is equal to co sign in Worse 47/52 which equals 25.3 degrees.
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