(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Besides its weight, a 4.20 kg object is subjected to one other constant force. The object starts from rest and in 1.20 s experiences a displacement of (3.80i hat - 3.30j hat) m, where the direction of j hat is the upward vertical direction. Determine the other force.

2. Relevant equations

F = ma; F =[tex]\sqrt{F_{x}^{2}+F_{y}^{2}}[/tex]

3. The attempt at a solution

okay so i know that the weight is pulling down on the object at 41.16 N. So i need to find the direction and magnitude of a force countering it, so that the resultant vector will = 5.03, however i'm confused about what role 1.20 s has to do with the problem. any help?

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# Two forces on a point particle

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