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ivanzaharov [21]
3 years ago
7

Hooke's law describes a certain light spring of unstretched length 33.6 cm. when one end is attached to the top of a doorframe a

nd a 6.89 kg object is hung from the other end, the length of the spring is 43.2 cm.
Physics
1 answer:
masha68 [24]3 years ago
4 0
Missing question: "What is the spring's constant?"

Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
F=mg=(6.89 kg)(9.81 m/s^2)=67.6 N
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
\Delta x=43.2 cm-33.6 cm=9.6 cm=0.096 m
And by using Hook's law, we can find the constant of the spring:
k= \frac{F}{\Delta x}= \frac{67.6 N}{0.096 m}=704.2 N/m
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The force F acting on a particle constrained to move along the x - axis is given by the fucntion F ( x ) = a x ( b − c x 2 ) [ a
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Hope this Helps!!!

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