How Does a Grapefruit Affect Spring Compression?

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The discussion centers on calculating the net force and acceleration of a grapefruit attached to a spring when it is 10 cm below the unstretched position. The correct net force is derived from the equation Fnet = mg - kx, where the mass is 0.289 kg, leading to a net force of approximately 0.962 N downward and an acceleration of 3.33 m/s². Participants clarify that the mass should be converted to kilograms for accurate calculations. The conversation also touches on determining the equilibrium position where the spring will come to rest, although specific calculations for that part are not detailed. The focus remains on understanding the forces acting on the grapefruit and correcting the initial mass used in the calculations.
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question: a grapefruit of mass 289g is attached to an unstretched vertical spring of force constant 18.7 N/m, and is allowed to fall.

a) determine the net force and the acceleration on the grapefruit when it is 10.0 cm below the unstretched position and moving downward.

b)air resistance will cause the grapefruit to come to rest at some equilibrium position. How far will the spring be stretched?


Relevant equations: Fspring=kx Espring=1/2kx^2



My attempt: Fnet=mg-kx I am getting 26.452 but the answer for part a) is 0.962 down and 3.33ms/^2
 
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Hi Physicsdb. Welcome to Physics Forums.

Show us your work for how you arrived at the number 26.452. Also, what are its units?
 
i figured to find the net force it would be the force of gravity minus the force of the string,
Fnet=Fg-Fs
Fnet=mg-kx
Fnet=2.89kg(9.8m/s^2)-(18.7N/m)(0.1m)
Fnet=28.322-1.87
Fnet=26.452 N
 
physicsdb said:
i figured to find the net force it would be the force of gravity minus the force of the string,
Fnet=Fg-Fs
Fnet=mg-kx
Fnet=2.89kg(9.8m/s^2)-(18.7N/m)(0.1m)
Fnet=28.322-1.87
Fnet=26.452 N
That should be 0.289 kg.
 

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