Question

How to solve this problem ?

: A big olive (m = 0.14 kg) lies at the origin of an xy coordinate system, and a big Brazil nut (M = 0.48 kg) lies at the point (0.79, 2.2) m. At t = 0, a force Upper F Overscript right-arrow EndScripts Subscript 0 Baseline equals left-parenthesis 2.4 i Overscript ̂ EndScripts plus 4.1 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 3.7 i Overscript ̂ EndScripts minus 1.4 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 = 4.5 s, with respect to its position at t = 0?

Answer 1

Net force F = Fo + Fn = (2.4 i + 4.1 j) + (-3.7 i - 1.4 j)
F = (-1.3 i + 5.5 j)N
Acceleration of center of mass a = F/(M+m)
= (-1.3 i + 5.5 j)/(0.48 + 0.14)
a = (-2.096 i + 8.87 j)
Time t = 4.5 s
Displacement of center of mass S = vo * t + (1/2)at^2
= 0 + [(1/2)*(-2.096 i +8.87 j)] * 4.5^2
= (- 21.222 i + 89.808 j) meter