posted by Tina .
A mass of 5.6 kg lies at the top of a frictionless incline. It is inclined at an angle such that the normal force on the mass is 30 Newtons. If the length of the incline is 58 meters and the mass is originally at rest at the top of the incline and then released, how long does it take for the mass to reach the bottom of the incline in seconds?
W = mg = 5.6kg * 9.8N/kg = 54.88N. = Weight of the mass.
Fm = (54.88N,Adeg) = Force of the mass.
Fv = 54.88cosA = 30N.
cosA = 30 / 54.88 = 0.5466,
A = 56.86 deg.
h = 58sin56.86 = 48.57m. = height of the incline.
h - Vo*t + 0.5g*t^2 = 48.57m,
0 + 4.9t^2 = 48.57,
t^2 = 9.91,
t(dn) = 3.15s.