Answered

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A 50-cm-long spring is suspended from the ceiling. A 210 g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 16 cm before coming to rest at its lowest point. It then continues to oscillate vertically.

Sagot :

The value of spring constant (k) is 25 N/m.

Conservation of energy, potential energy of a stretched spring, gravitational potential energy, frequency of oscillation in a spring coupled to a mass, and variation of acceleration due to gravity with altitude are the principles necessary to answer the given problem.

To begin, use energy conservation to equal change in gravitational potential energy on mass attached to spring and elastic potential energy of stretched spring. Then, in the equation, substitute the specified numbers and solve for the spring constant. Then, calculate the amplitude by taking half of the value of spring stretching from equilibrium, and the frequency of oscillation by using the calculation for frequency of oscillation in a stretched string in terms of mass and spring constant.

Formula apply

U=mgh

U=1/2kx^2

putting the value of K ,x ,mass and g then

k is 25N/m

To learn more about spring constant from the given link:

https://brainly.com/question/14159361

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