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anonymous
 5 years ago
A 50.0 kg wagon is towed up a hill inclined at a constant angle. The tow rope is parallel to the incline and exerts a force of 150 N. Assume that the wagon starts from rest at the bottom of the hill, neglect friction. What is the Maximum angle that the slope of the hill can be so that the wagon to be moving at 10.0 m/s at 60.0m up the hill?
anonymous
 5 years ago
A 50.0 kg wagon is towed up a hill inclined at a constant angle. The tow rope is parallel to the incline and exerts a force of 150 N. Assume that the wagon starts from rest at the bottom of the hill, neglect friction. What is the Maximum angle that the slope of the hill can be so that the wagon to be moving at 10.0 m/s at 60.0m up the hill?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is a question for physics...Had this kind of tests in my physics before.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so Ms. Bio, what is the answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes it it. and i dont know how to do it

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.2ok i think i figured it out m = 50, F = 150, A = angle F = ma Need to find net Force negative force parallel to ramp caused by gravity is mg*sin(A) g = 9.8 NetForce = 150  490sin(A) a = F/m = netForce/50 a = 3  9.8sin(A) velocity in antiderivative of acceleration w/ respect to time v = (39.8sin(A))t position or distance is antiderivative of velocity x = (39.8sin(A))/2 t^2 Now find t to satisfy 2 conditions, v=10 and x=60 from velocity equation t = 10/(39.8sin(A)) substitute this into distance equation 60 = (39.8sin(A))/2 * (10/(39.8sin(A)))*t 60 = 5t 12 = t replace t with 12 find A 12 = 10/(39.8sin(A)) 39.8sin(A) = 10/12 sin(A) = (10/12  3)/9.8 sin(A) = .2211 A = sin^1(.2211) A = 12.77 degrees
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