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A relaxed spring of length 0. 13 m stands vertically on the floor; its stiffness is 1180 N/m. You release a block of mass 0. 5 kg from rest, with the bottom of the block 0. 7 m above the floor and straight above the spring. How long is the spring when the block comes momentarily to rest on the compressed spring?.

Respuesta :

The length of spring when the block comes momentarily to rest on the compressed spring will be 0.054 m.The length of the spring by the letter x.

What is the potential energy of the spring?

The energy is stored in the spring when it is stretched or compressed by some length. It is the product of mass, gravity and distance compressed or stretched. Mathrmatically it is given by;

PE=mgh

The given data in the problem is ;

m is the mass of the block is 0.5 kg height,

h is the height is released is 0.7 m

x initial length of the spring = 0.13 m.

K is the force constant of the spring = 1180 N/m.

By the law of conservation of energy,

The potential energy of the spring gets converted

[tex]\rm mgh=\frac{1}{2}KL^2 \\\\ \rm L= \sqrt{\frac{2mgh}{K} } \\\\ \rm L= \sqrt{\frac{2\times 0.5\times9.81 \times 0.7}{1180} \\\\ \\\\[/tex]

[tex]\rm L= 0.054m[/tex]

Hence the length of spring when the block comes momentarily to rest on the compressed spring will be 0.054 m

To learn more about the potential energy of the spring refer to the link;

https://brainly.com/question/2730954