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) i + (4.00 m) j - (3.00 m) k , its velocity is v = - (2.07m/s) i + (6.41 m/s) j + (5.42 m/s) k and it is subject to a force F = (9.89 N) i - (3.87 N) j + (4.34 N) k . Find the acceleration of the object ((a), (b) and (c) for i , j and k components respectively), the angular momentum of the object about the origin ((d), (e) and (f) for i , j and k components respectively), the torque about the origin acting on the object ((g), (h) and (i) for i , j and k components respectively), and (j) the angle between the velocity of the object and the force acting on the object.

Explanation / Answer

acceleration=force/mass

=4.945 i -1.935 j +2.17 k

part a:

i component=4.945 m/s^2

part b:

j component=-1.935 m/s^2

part c:

k component=2.17 m/s^2

part d:

angular momentum=cross product of position vector and momentum vector

=cross product of (2 i + 4j -3k) and 2*(-2.07 i +6.41 j +5.42 k)

=81.82 i -9.26 j +42.2 k

part d:

i component=81.82

part e: j component=-9.26

part f: k component=42.2

torque=cross product of position vector and force vector

=cross product of (2 i +4 j -3k) and (9.89 i -3.87 j +4.34 k)

=5.75 i -38.35 j -47.3 k

part g:

i component of torque=5.75 N.m

part h:

j component of torque=-38.35 Nm

part i:

k component of torque=-47.3 N.m

part j:

angle between velocity and force

=arccos(dot product of two vectors/(product of magnitude of two vectors))

=arccos((2*9.89-4*3.87-3*4.34)/(sqrt(2^2+4^2+3^2)*sqrt(9.89^2+3.87^2+4.34^2)))

=98.114 degrees

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