At the instant the displacement of a 2.00 kg object relative to the origin is d

Question

At the instant the displacement of a 2.00 kg object relative to the origin is d = (2.00 m) j + (4.00 m) (3.00 m) k , its velocity is v =-(9.10 m/s) j + (3.51 m/s) j + (8.46 m/s) k and it is subject to a force F (3.80 N) j-(3.31 N) j + (3.60 N) k . Find the acceleration of the object ((a), (b) and (c) for j , j and k components respectively), the angular momentum of the object about the origin ((d), (e) and (f) for , , and K components respectively), the torque about the origin acting on the object ((g), (h) and (i) for j and k components respectively), and j) the angle between the velocity of the object and the force acting on the object.

Explanation / Answer

a)

ax = Fx/m = 3.80/2 = 1.9 m/s2

b)

ay = Fy/m = - 3.31/2 = - 1.66 m/s2

c)

az = Fz/m = 3.60/2 = 1.8 m/s2

d)

angular momentum is given as

L = m (r x v) = (2) ((2i + 4j - 3 k) x (- 9.1 i + 3.51 j + 8.46 k))

L = 2 (44.4 i + 10.4 j + 43.4 k)

L = 88.8 i + 20.8 j + 86.8 k

Lx = 88.8

Ly = 20.8

Lz = 86.8

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