10-28-2013, 07:33 PM
I'm not sure that problem makes sense. First, is the object starting from rest or does it have an initial velocity? Are we assuming it's traveling horizontally across a friction-less surface or is some other kind of motion involved? Is this object not on Earth because a normal force of 65.5 N would mean that the object weighs roughly 6.7 kg under Earth's gravity. And if we're assuming that the 5.63 m/s^2 is actually acceleration due to gravity, it still gives an 11.6 kg object.
All that being said, the formula you're likely going to be using is x = (1/2)at^2, so, plugging in, you get (1/2)(5.63)(5)^2 = 2.815*25 = 70.4 m, assuming that the object starts from rest. If you are not starting from rest, the formula becomes vt+(1/2)at^2 where v = initial velocity, a is still 5.63 m/s^2, and t is 5 seconds meaning your answer would be 5v + 70.4 m. Normal force and mass are irrelevant in this case, given that there is no friction.
All that being said, the formula you're likely going to be using is x = (1/2)at^2, so, plugging in, you get (1/2)(5.63)(5)^2 = 2.815*25 = 70.4 m, assuming that the object starts from rest. If you are not starting from rest, the formula becomes vt+(1/2)at^2 where v = initial velocity, a is still 5.63 m/s^2, and t is 5 seconds meaning your answer would be 5v + 70.4 m. Normal force and mass are irrelevant in this case, given that there is no friction.