contestada

A block of mass m = 0.23 kg is set against a spring with a spring constant of k1 = 553 N/m which has been compressed by a distance of 0.1 m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k2 = 370 N/m. How far d2, in meters, will the second spring compress when the block runs into it?

Respuesta :

Answer:

[tex]X_2 = 0.122m[/tex]

Explanation:

Using the law of the convervation of energy E:

[tex]E_i = E_f[/tex]

so:

[tex]\frac{1}{2}K_1X_1^2 = \frac{1}{2}K_2X_2^2[/tex]

where [tex]K_1[/tex]is the constant of the spring 1, [tex]X_1[/tex] the compressed of the first spring, [tex]K_2[/tex] is the constant of the second spring and [tex]X_2[/tex] is the compressed of the second spring.

Replacing values, we get:

[tex]\frac{1}{2}(553N/m)(0.1m)^2 = \frac{1}{2}(370N/m)X_2^2[/tex]

Solving for [tex]X_2[/tex]:

[tex]X_2 = 0.122m[/tex]