At the instant the displacement of a 2.00 kg object relative to the origin is d
ID: 2239785 • Letter: A
Question
At the instant the displacement of a 2.00 kg object relative to the origin is d = (2.00 m) + (4.00 m) - (3.00 m) , its velocity is v = - (6.20 m/s) + (3.00 m/s) + (2.90 m/s) , and it is subject to a force F = (6.40 N) - (8.90 N) + (3.70 N) . (a) Find the acceleration of the object. (in i,j, and k) (b) Find the angular momentum of the object about the origin. (in i, j, and k) (c) Find the torque about the origin acting on the object. (in i, j, and k) (d) Find the angle between the velocity of the object and the force acting on the object.Explanation / Answer
a) acclereration = F/m = (6.40i - 8.90j+3.70k)/2 =3.20i-4.45j+1.85k m/s^2 b) Angular Momentum = m(r X v) =2(20.6i-12.8j+30.8k) =41.2l-25.6j+61.6k c)Torque =rXF =41.5i-34.1j+43.4k Nm
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.