Asked by Brena
Water skiers often ride to one side of the center line of a boat, as shown in the figure . In this case, the ski boat is traveling at 15 (m/s} and the tension in the rope is 80 N.
If the boat does 3600 J of work on the skier in 51.3 m, what is the angle \theta between the tow rope and the center line of the boat?
Someone please help been trying for hours...
If the boat does 3600 J of work on the skier in 51.3 m, what is the angle \theta between the tow rope and the center line of the boat?
Someone please help been trying for hours...
Answers
Answered by
drwls
The component of the rope tension along the direction of motion is
(Work)/(distance pulled) = 3600/51.3 = 70.2 N
But you also know that
70.2 = (rope tension)x cos theta
Therefore
sin theta = 0.877
theta = 28.7 degrees
(Work)/(distance pulled) = 3600/51.3 = 70.2 N
But you also know that
70.2 = (rope tension)x cos theta
Therefore
sin theta = 0.877
theta = 28.7 degrees
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