An ornament of mass 40.0 g is attached to a vertical ideal spring with a force constant (spring constant) of 20.0 N/m. The ornament is then lowered very slowly stops stretching. How much does the spring stretch?A) 0.00200 m.B) 0.0196 m.C) 0.0816 m.D) 0.800 m.E) 0.200 m.

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Answer:

Option A is the correct answer.

Explanation:

We have force is the product of spring constant and distance stretched by spring.

             Force = Spring constant x Distance stretched by spring

                F = kx

Force = Mass x acceleration

Mass = 40 g = 0.04 kg

Acceleration = 9.81 m/s²

Force = Mass x acceleration = 0.3924 N

              F = kx

             0.3924 = 20x

                x = 0.0196 m

Option A is the correct answer.

The stretched length of the spring is 0.0196 m. Option B shows the correct stretched length of the spring.

What is the spring constant?

The spring constant is defined as the applied force if the displacement in the spring is unity.

Given that the mass m of the ornament is 40 g and the spring constant is 20.0 N/m.

The spring is attached in the vertical direction, hence the force acted on the ornament is given below.

[tex]F = mg[/tex]

Where g is the gravitational acceleration.

The force on the spring is given as below.

[tex]F = kx[/tex]

Where k is the spring constant and x is the stretched length of the spring.

When equating both the forces, we get,

[tex]mg = kx[/tex]

[tex]0.040 \times 9.8 = 20.0 \times x[/tex]

[tex]x = 0.0196 \;\rm m[/tex]

Hence we can conclude that the stretched length of the spring is 0.0196 m. Option B shows the correct stretched length of the spring.

To know more about the spring constant, follow the link given below.

https://brainly.com/question/4291098.