A big olive m :0 10 kg) lies at the origin of an xy coordinate system, and a big

Question

A big olive m :0 10 kg) lies at the origin of an xy coordinate system, and a big Brazil nut Ms 0.48 kg lies at the pont 0.77 2.4) m At t = 0, a force 0 3.7, + 1.4 j N begins to act on the olive, and a force F N 1.8, 3.5 N begins to act on the nut, what is the (a) x and (b) y displacement of the center of mass of the olive-nut system at t 7.0 s, with respect to its position at t = 0? (a) Number (b) Number Click if you Units Units Tm The number of significant digits is set to 2; the tolerance is +-1 in the 2nd significant digit

Explanation / Answer

Net force F = Fo + Fn = (3.7 i + 1.4 j) + (-1.8 i – 3.5 j)

F = (1.9 i – 2.1 j)N

Acceleration of center of mass a = F/(M+m)

= (1.9 i – 2.1 j)/(0.1 + 0.48)

a =(1.9 i – 2.1 j)/0.58

Initial velocity of CM vo = 0

Time t = 7.0 s

Displacement of center of mass S = vo * t + (1/2)at^2

= 0 + [(1/2)* (1.9 i – 2.1 j)/0.58] * 7.0^2

=[(1.9 i – 2.1 j)/1.16] * 49

= (80.26 i – 88.71j) meter

Thus:

(a) x = 80.26m

(b) y = -88.71m

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