contestada

2)
The staples inside a stapler are kept in place by a spring with a relaxed length of 0.115 m. If the
spring constant is 51.0 N/m, how much elastic potential energy is stored in the spring when its
length is 0.150 m?

Respuesta :

opudodennis

Answer:

0.0312J

Explanation:

Let x be the distance the staple moves:

[tex]x=0.150m-0.115m=0.035m[/tex]

And spring constant is [tex]k=51.0N/m[/tex]

[tex]PE=0.5kx^2\\=0.5\times 51.0\times 0.035^2\\\\=0.312[/tex]

Hence, the potential energy is 0.0312J

Cricetus

The stored potential energy will be "0.312 J".

According to the question.

Length,

L = 0.115 m

Spring constant,

k = 51.0 N/m

Now,

→ [tex]P.E_{elastic} = \frac{1}{2} kx^2[/tex]

                  [tex]= \frac{1}{2}\times 51(0.150-0.115)^2[/tex]

                  [tex]= \frac{0.01225\times 51}{2}[/tex]

                  [tex]= 0.312 \ J[/tex]

Thus the above answer is correct.

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https://brainly.com/question/14172051

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