Question Part Points Submissions Used Besides the gravitational force, a 2.75-kg

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Besides the gravitational force, a 2.75-kg object is subjected to one other constant force. The object starts from rest and in 1.20 s experiences a displacement of (5.50 ? 3.30) m, where the direction of is the upward vertical direction. Determine the other force.

Explanation / Answer

let the displacement vector be d = dx (i) + dy(j) = 3.80 (i) - 3.30 (j) ----------- (1) resolving the motion in x & y directions (vertical) dx = 0 (rest) + 0.5 [ax] t^2 = 3.80 dy = 0 (rest) + 0.5 [ay] t^2 = - 3.30 where ax and ay are components of acceleration produced by respective forces let >>>>>> F = Fx (i) + Fy (j) ax = Fx/m ay = [Fy - mg]/m >>>>>> weight acting vertically down 0.5 [ax] t^2 = 3.80 0.5 [ay] t^2 = - 3.30 -------------------------------- [ax] = 3.80*2/1.2*1.2 = Fx/m [ay] = - 3.30*2/1.2*1.2 = [Fy - mg]/m =Fy/m - g Fx = 20.583 Fy = 20.345 --------------------------------- F = 20.583 (i) + 20.345 (j) this is the force applied

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