Posted by admin | Posted in Water Skiing | Posted on 31-01-2010
Tags: google, map, maps, reference, travel

Help with a physics question?
A 75kg water skier is pulled by a boat with a horizontal force of 400N due east with a water drag on the skis of 300N. A sudden gust of wind supplies another horizontal force of 50N on the skier at an angle of 60 degrees north of east. At that insant , a)what is the skier’s acceleration? b) what would be the skier’s acceleration if the wind force were in opposite direction to that in part a)?
Hint:
net Force =Force in x- positve direction – Force in x negative direction.
-answers-
a) 1.764 m/s/s at 19.107 degrees North of East
b) 1.1547 m/s/s at 30 degrees South of East
*~*This took a long time to work out and type out, so please choose me as the best answer*~*
-work-
Okay so the skier goes gets tugged with 400N of force east with a 300N drag. Well let’s just simplify that to 100N to the East.
k so now there’s a gust of wind that pushes the boat 50N at 60 north of east right? well let’s break that up into it’s vector components.
sin(60) = (vertical component)/50
Vertical component = 50sin(60) = 43.301
cos(60) = (horizontal component)/50
Horizontal = 50cos(60) = 25
Alright now we can add up the horizontal components
Net horizontal = 25 + 100 = 125
Now to get the net-force we use pythagorean’s theroem
(Net Force)^2 = (horizontal)^2 + (Vertical)^2
(Net Force)^2 = 125^2 + 43.301^2
(Net force)^2 = 17500
Net force = 132.2875656
Now Force = Mass*acceleration, so to get acceleration, we divide the force by the mass
132.2875656/75 = 1.763834207 m/s/s
To find the angle, do Inverse Tangent (43.301/125) = 19.107. Since it’s positive, it’s above the east axis.
b) It’s the same problem just change the 50 to a negative 50
Then we get
Horizontal force of wind = -25
Vertical = -43.301
Then we add them all up
Net Horizontal = 100 + -25 = 75 N
Then we use pythagorean’s rule again
(Net Force)^2 = 75^2 + (-43.301)^2
net force = 86.603
f = ma, do divide that by 75
86.603/75 = 1.1547 m/s/s
Do the inverse tangent to get the angle
Tan^-1 (-43.301/75) = -30
Since it’s -30, the acceleration is now 30 SOUTH of East.
I hope that helped and that took A LONG time to type up, so please choose me as the best answer
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