Asked by Jon
A mass of 11 kg is being pushed up a frictionless incline with a constant force of 35 Newtons directed parallel to the incline, up the incline. At the top of the incline the mass is moving at 29 m/s up the incline. If the angle of inclination is 59 degrees and the height (not length) of the incline is 23 meters, what was the magnitude of the mass's velocity at the bottom of the incline in m/s?
Answers
Answered by
Henry
Vf = (29m/s,59deg).
Yf = ver. = 29sin59 = 24.85m/s. = Vertical component.
Yf^2 = Yo^2 + 2g*d,
Yo^2 = Yf^2 - 2g*d,
Yo^2 = (24.85)^2 - 19.6*23 = 166.7,
Yo = 12.9m/s. = Ver. component of initial velocity.
Vo = Yo / sin59=12.9 / sin59=15.1m/s. =
Initial velocity.
Yf = ver. = 29sin59 = 24.85m/s. = Vertical component.
Yf^2 = Yo^2 + 2g*d,
Yo^2 = Yf^2 - 2g*d,
Yo^2 = (24.85)^2 - 19.6*23 = 166.7,
Yo = 12.9m/s. = Ver. component of initial velocity.
Vo = Yo / sin59=12.9 / sin59=15.1m/s. =
Initial velocity.
Answered by
Tina
A mass of 18 kg is being pushed up a frictionless incline with a constant force of 20 Newtons directed parallel to the incline, up the incline. At the top of the incline the mass is moving at 23 m/s up the incline. If the angle of inclination is 58 degrees and the height (not length) of the incline is 24 meters, what was the magnitude of the mass's velocity at the bottom of the incline in m/s?
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