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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this is a question for physics...Had this kind of tests in my physics before.
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ok 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 anti-derivative of acceleration w/ respect to time
v = (3-9.8sin(A))t
position or distance is anti-derivative of velocity
x = (3-9.8sin(A))/2 t^2
Now find t to satisfy 2 conditions, v=10 and x=60
from velocity equation
t = 10/(3-9.8sin(A))
substitute this into distance equation
60 = (3-9.8sin(A))/2 * (10/(3-9.8sin(A)))*t
60 = 5t
12 = t
replace t with 12 find A
12 = 10/(3-9.8sin(A))
3-9.8sin(A) = 10/12
sin(A) = (10/12 - 3)/-9.8
sin(A) = .2211
A = sin^-1(.2211)
A = 12.77 degrees