What Is the Other Force Acting on the Object?

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SUMMARY

The discussion focuses on determining the additional force acting on a 4.20 kg object subjected to gravity and another constant force. The object experiences a displacement of (3.80i - 3.30j) m in 1.20 seconds, with the weight calculated at 41.16 N. The participant attempts to calculate the force components using the equations of motion, specifically F = ma and the kinematic equation for displacement. However, discrepancies arise in the calculations for the force components, particularly for Fy, which resulted in an incorrect value of 376.275 N, indicating a significant error in the approach.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Familiarity with kinematic equations for uniformly accelerated motion
  • Knowledge of vector decomposition in physics
  • Basic grasp of gravitational force calculations
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  • Review the application of kinematic equations in two dimensions
  • Study vector decomposition techniques for force analysis
  • Learn about the role of time in calculating acceleration and force
  • Practice solving problems involving multiple forces acting on an object
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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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