A mass of 6 kg is resting at the bottom of a frictionless incline with a length

Question

A mass of 6 kg is resting at the bottom of a frictionless incline with a length of 3.1 meters and an inclination angle of 37 degrees. Force 1 has a magnitude of 1058 Newtons is applied to it directly up the incline, and force 2 is applied down the incline at an angle of 17 degrees below the direction parallel to the incline (see sketch). When these forces are applied, the normal force acting on the mass is twice amount of its weight. When the mass reaches the top of the incline, it becomes a projectile where only gravity acts on it. What maximum height above the ground (bottom of the incline) does the mass reach in meters?

Explanation / Answer

given mass m = 6 kg
frictionless incline
l = 3.1 m
theta = 37 deg

F1 = 1058 N
F2 = ?, phi = 17 deg

also normal force N = 2mg
now
from force balance

N = mg*cos(theta) + F2*sin(phi)
F2 = [2mg - mgcos(theta)]/sin(phi) = 6*9.81(2 - cos(37))/sin(17)
F2 = 241.8575831258 N

hence
acceleraiton = a
m*a = F1 - mg*sin(theta) - F2*cos(phi)
a = F1/m - g*sin(theta) - F2*cos(phi)/m
a = 131.88126 m/s/s
2*a*l = v^2
v = 28.594822335857761157880790796534 m/s

2*g*hmax = v^2*sin^2(theta)
hmax = 15.09391332527 m above the top end of the linclien

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