Question
A 4.2 kg object is subjected to two forces,
F~
1 = (2.1 N) ˆı + (−1.7 N) ˆ and F~
2 =
(3.2 N) ˆı + (−11.2 N) ˆ. The object is at
rest at the origin at time t = 0.
What is the magnitude of the object’s acceleration?
F~
1 = (2.1 N) ˆı + (−1.7 N) ˆ and F~
2 =
(3.2 N) ˆı + (−11.2 N) ˆ. The object is at
rest at the origin at time t = 0.
What is the magnitude of the object’s acceleration?
Answers
Damon
Fx = 2.1 +3.2 = 5.3
Fy = -1.7 -11.2 = - 12.9
|F| = sqrt ( 5.3^2 + 12.9^2)
|a| = |F|/4.2 meters/second^2
Fy = -1.7 -11.2 = - 12.9
|F| = sqrt ( 5.3^2 + 12.9^2)
|a| = |F|/4.2 meters/second^2