Acceleration of 3.8kg Object: 110.58 m/s2, 28.66°

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The discussion focuses on calculating the acceleration of a 3.8 kg object subjected to two forces: 40 N directed east and 75 N at a 45-degree angle. The resultant force is determined using the Pythagorean theorem, yielding a magnitude of 97.03 N and an acceleration of 110.58 m/s² at an angle of 28.66 degrees relative to the x-axis. The calculations involve resolving the forces into their x and y components and applying trigonometric functions to find the resultant vector.

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Only two forces act on an object (mass= 3.8 kg). One force is pointing east with 40 N. The other has a force of 75.0 N at a 45 degree angle. Find the magnitude and direction (relative to the x-axis) of the acceleration of the object.



pythagorean theorum
tan O= opp/adj



x y
53.03 53.03 < ---- found using 75sin45
40 0

= 97.03, 53.03 <--- resultant

pythagorean theorum with resultants = 110.58m/s
angle from tangent= 28.66 degrees
 
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Hi wallace13,

wallace13 said:
Only two forces act on an object (mass= 3.8 kg). One force is pointing east with 40 N. The other has a force of 75.0 N at a 45 degree angle. Find the magnitude and direction (relative to the x-axis) of the acceleration of the object.



pythagorean theorum
tan O= opp/adj



x y
53.03 53.03 < ---- found using 75sin45
40 0

= 97.03, 53.03 <--- resultant

I believe that the 97.03 is incorrect.
 

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