The force F = (20, -30) N acts on a mass of 8 kg. At time t = 0 s, the mass is i

Question

The force F = (20, -30) N acts on a mass of 8 kg. At time t = 0 s, the mass is initially at rest. In which quadrant (1-4) is the force? Submit Answer Tries 0/3 What is the angle (in degrees) of the force, measured counter-clockwise from thex-axis? Submit Answer Tries 0/3 What is the magnitude of the force (in N)? Submit Answer Tries 0/3 What are the (x,y) components of the acceleration vector (in m/s2)? Subimit Answer Tries o/3 What are the (x,y) components of the velocity vector (in m/s) after 0.6 seconds? Submit Answer Tries 0/3

Explanation / Answer

According to the given problem,

1.)The force is in 4th quadrant as x is positve and y is negative.

2.) Using the realtion,

Tan-1[y/x] = 360+ Tan[-30/20]

= 303.7o counter clockwise from x-axis.

3.)The magnitude of force,

= 202 + 302

= 36.1N.

4.) The accelration is,

F = ma

a = F/m

a = [20,-30]/8

a = [2.5, -3.75].

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