Phy 231
Your 'cq_1_17.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I drew the cart midway down the incline, so part of the incline is in the first quadrant and part is in the third. The incline makes a 30 degree angle with the positive x-axis, so since the gravitational force is acts vertically downward at 5kg(-9.8ms^2) = 49 N, it'll be on the y axis with two components, making a 30 deg + 90 deg = 120 deg angle with the positive x-axis, and the parallel gforce will, of course, just be parallel at 30 deg. So, the perpendicular component will be 49 cos(120 deg) = -24.5 N, and the parallel component will be 49sin(30 deg) = 24.5 N.
So the normal force would be 24.5 N in the upward direction to balance the -24.5N in the downward direction.
Since that's balanced,
Fnet = 24.5, and 24.5 N = 5kg * a, and a = 4.9m/s^2
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15-20 min
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Again good, and I recognize the question from your well-stated and detailed response. Do remember to insert your answers into a copy of the question on future cq's.
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