A woman weighs 120 lb. Determine (a) her weight in newtons (N) and (b) her mass in kilograms (kg). My book tells me that 1 N is about .25 of a pound. So, that makes her weight 480 N. F = ma, I know that, but I'm not exactly sure how to get accleration, unless I'm assuming it's on earth at -9.8 m/s^2. Another problem: Besides its weight, a 2.80-kg object is subjected to one other constant force. The object starts from rest and in 1.20 s experiences a displacement of (4.20i - 3.30j) m, where the direction of j is the upward vertical direction. Determine the other force. What I do not get is that the value of j being upward if it has a negative sign, but, well. :S I know that I can perhaps get the velocity by dividing the components by 1.20, or taking the square root of the components squared and addd together (the magnitude) and then divide that value by 1.20, but again, that's just the velocity. Taking its derivative would be ...zero. Okay, so then, would the force be zero? If F = (2.80)(0)? Yet another problem: Three forces acting on an object are given by F1 = (-2.00i + 2.00j)N, F2 = (5.00i - 3.00j) N, and F3 = (-45.0i) N. Thge object experiences an acceleration of magnitude 3.75m/s^2. (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0s? (d) What are the velocity components of the object after 10.0s? Fx = -42 Fy = -1 (a) I'm not sure if this method is correct, however, for getting an angle using only the forces, when the problem asks for an angle from the acceleration: tan[angle] = (-1/-42) = 1.364, about. (b) With a = 3.75 and the net force as [-42, -1], the mass is about 11.2 kg. (c) v[final] = v[inital] + at = 0 + (3.75)(10) = 37.5 m/s. (d) If my angle is correct, I'm hoping to use trig to finish this portion of the problem: v[x] = 37.5cos1.364 v[y] = 37.5sin1.364 I'm just wondering, for the last problem above, if my train of thought was correct. Thank you!