What Is the Other Force Acting on the Object?

AI Thread Summary
A 4.20 kg object experiences a constant force in addition to its weight, which is 41.16 N downward. The object moves 3.80 m in the horizontal direction and -3.30 m vertically in 1.20 seconds, indicating a need to calculate the net force acting on it. The time duration is crucial for determining the acceleration components using kinematic equations. The calculations for the horizontal and vertical forces must be set up correctly, as errors in the equations can lead to significant discrepancies in the results. Properly analyzing the forces and their components is essential for finding the correct answer.
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Homework Statement



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.

Homework Equations



F = ma; F =\sqrt{F<sub>x</sub><sup>2</sup>+F<sub>y</sub><sup>2</sup>}

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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You can break this problem down into its component constituents... the force due to gravity acts only in the j direction, so when this particle is displaced 3.80m in the i direction, in 1.20s, for which the acceleration is constant (because force is constant), you have:

3.80 = \frac{1}{2}a_x(1.20)^2

since the object starts at rest (this is a suvat equation). Then you have:

F_x = ma_x

Under the force of gravity alone, the object will travel some distance in j direction but has in fact traveled -3.30m in that direction... find out the difference and see what extra distance you need and then set up another equation the same as the one i have in x, with

s = 1/2a_y(1.20)^2

where s is this 'extra' distance.

The importance of the 1.20s then, is in calculating the component accelerations (and hence the forces) of the particle.
 
okay i did all that, and for the Fx, i still got the wrong answer. i got 23.76, but the program I'm using said i was within 10% of the answer. for the Fy, however, it said i was off by orders of magnitude. Here is what i did, maybe you can show me where i went wrong:

Yf = m(-g)t2 + s
then i plugged in numbers, solved for x, and put that back into that
s = 1/2ay(1.20)2,
found ay and then
plugged that into Fy = may
and got a final answer of 376.275, but it said i was off by orders of magnitude.
 
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