FO
1.0 A 0.250 kg block attached to a light spring oscillates on a frictionless, horizontal table. The
oscillation amplitude is A = 0.125 m and the block moves at 3.00 m/s as it passes through
equilibrium at x =
<= 0.
(a) Find the spring constant

Respuesta :

The spring constant, k, will be equal to 18 N/m. The spring constant is defined as the force-to-displaced-length ratio.

What is the spring constant?

Spring constant is defined as the ratio of force per unit displaced length.

Given data;

Mass of  block,m=0.250 k

Amplitude,A = 0.125 m

Velocity,V=3.00 m/

Spring constant, K=?

The spring constant formula when the spring oscillates at the specified amplitude A:

[tex]\rm k= \frac{v_{max}^2 m}{A} \\\\ k =\frac{3.0 m/s^2 \times 0.250 \ kg}{0.125 \ m} \\\\ k=18 \ N/m[/tex]

Hence, the spring constant, k will be 18 N/m

To learn more about the spring constant refer:

https://brainly.com/question/4291098

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