A big olive (m = 0.097 kg) lies at the origin of an xy coordinate system, and a

Question

A big olive (m = 0.097 kg) lies at the origin of an xy coordinate system, and a big Brazil nut (M = 0.83 kg) lies at the point (0.74, 3.5) m. At t = 0, a force Upper F Overscript right-arrow EndScripts Subscript 0 Baseline equals left-parenthesis 2.1 i Overscript EndScripts plus 1.0 j Overscript EndScripts right-parenthesis N begins to act on the olive, and a force Upper F Overscript right-arrow EndScripts Subscript Upper N Baseline equals left-parenthesis negative 3.1 i Overscript EndScripts minus 1.6 j Overscript EndScripts right-parenthesis 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 = 7.3 s, with respect to its position at t = 0?

Explanation / Answer

Net force F = Fo + Fn = (2.1 i + 1.0 j) + (-3.1 i - 1.6 j)
F = (-1.0 i +2.6 j)N
Acceleration of center of mass a = F/(M+m)
= (-1.0 i + 2.6 j)/(0.83 + 0.097)
a = (-1.0 i + 2.6 j)/0.927 m/s^2
Initial velocity of CM vo = 0
Time t = 7.3 s
Displacement of center of mass S = vo * t + (1/2)at^2
= 0 + [(1/2)*(-1.0 i + 2.6 j)/1.2] * 7.3^2
= [(-1.0 i + 5.0 j)/2.6] * 53.29
= -(53.29/2.6)i + (5 * 53.29/2.6)j
= (- 20.49 i + 102.48 j) meter

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