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 = -(9.35 m/s) j + (9.50 m/s) j + (3.75 m/s) k and it is subject to a force F = (8.27 N) j - (2.29 N) j + (7.54 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. (a) Number Units (b) Number Units (c) Number Units (d) Number Units (e) Number Units

Explanation / Answer

a = F/m

F = 8.27 i - 2.29 j + 7.54 k

m = 2kg

a = 4.135 i - 1.145j + 3.77k

part a )

4.135 m/s^2

b )

-1.145 j m/s^2

c )

3.77 ms/^2

L = r x p = m ( r x v)

r = 2i + 4j - 3k

v = -9.35 i + 9.50 j + 3.75 k

L = 87 i kg*m^2/s + 41.1 j kgm^2/s + 112.8 kg-m^2/s

d ) = 87 kg-m^2/s

e) = 41.1 kg-m^2/s

f ) = 112.8 kg-m^2/s

torque = r x F

t = 23.29 i - 39.89j - 37.66k

g ) = 23.29 N-m

h ) = -39.89 N-m

i) = -37.66 N-m

part j )

cos theta = v.F/(|v|*|F|)

|v| = sqrt(vx^2+vy^2+vz^2) = 13.8468

|F| = sqrt(Fx^2+Fy^2+Fz^2) = 11.42316

v.F = 3.9955

theta = cos^-1(v.F/(|v|*|F|))

theta = 88.55 degree

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