Question

Janet wants to find the spring constant of a given spring, so she hangs the spring vertically and attaches a 0.47 kg mass to the spring's other end. if the spring stretches 3.1 cm from its equilibrium position, what is the spring constant? 133 incorrect: your answer is incorrect. n/m

Answer 1

Answer:

  148.6 N/m

Explanation:

The 0.47 kg mass applies a force of ...

   F = ma = (0.47 kg)(9.8 m/s^2) = 4.606 N

which stretches the spring by 3.1 cm, or 0.031 m. That means the spring constant is ...

  (4.606 N)/(0.031 m) ≈ 148.6 N/m

_____

Additional comment

The supplied numbers have 2 significant figures. If the spring constant is rounded to 2 significant figures, it would be 150 N/m.

Answer 2

The force needed to extend or compress a spring is directly proportion to the applied stress in spring. The spring constant of a given spring is 148.71 N/m.

The Hook's law:

Within the elastic limit the force needed to extend or compress a spring is directly proportion to the applied stress in spring.

The force applied by the wight

$\bold {F = mg }$

Where,

m = mass = 0.47 kg

g - gravitational acceleration = $\bold {9.81 m /s^2}$

Put the values in the equation,

$\bold {F = 0.47 kg \times 9.81 m/s^2}\\\\\bold {F = 4.61}$

Spring constant can be calculated using the Hook's law,

$\bold {k = \dfrac {F}{x}}$

$\bold {k = \dfrac {4.61 N}{0.03 m}}\\\\\bold {k = 148.71N/m}$

Therefore, the spring constant of a given spring is 148.71 N/m.

To know more about Hook's law,

https://brainly.com/question/10991960