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A 263 g ball of clay falls onto a vertical spring.
The spring constant k = 2.52 N/cm.
The clay sticks to the spring and as the spring compresses a distance of 11.8 cm the clay momentarily comes to rest.
Find the maximum compression of the spring, in [cm], if the initial speed of the clay doubled?

Respuesta :

Answer:

d. According to the previous theorem:

(kh^2)/2-mgh-(m(2v)^2)/2=0,

126h^2-2.577h-8.236=0,

h=0.266 m.

Explanation:

The maximum compression of the spring should be 0.266 m.

Given that,

  • A 263 g ball of clay falls onto a vertical spring.
  • The spring constant k = 2.52 N/cm.
  • The clay sticks to the spring and as the spring compresses a distance of 11.8 cm the clay momentarily comes to rest.

Based on the above information, the calculation is as follows:

[tex](kh^2)\div 2-mgh-(m(2v)^2)\div 2=0\\\\126h^2-2.577h-8.236=0,[/tex]

h=0.266 m

Learn more: brainly.com/question/16911495

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