A 0.20kg mass attached to the end of a spring, is whirled in a vertical circle by a student. At some positions, the mass experiences a downward force of -62j N and a horizontal force of 38i N. Determine its acceleration in to this position.?
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Robert01
4 years ago
Fy= –62jN
Fx= 38iN
F= (38i – 62j)N {F= resultant force}
|F|= √((Fx)² + (Fy)²) = √((38)² + (–62)²) {|F|= modulus of the resultant force}
|F|= 72.72N
a= F/m or |F|/m
a= 72.72/0.20 or (38i – 62j)/0.20
= 363.6m/s² or (190i – 310j)m/s²
