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At the grocery store you place a pumpkin with a mass of 12.5 lb on the produce spring scale. The spring in the scale operates such that for each 5.7 lbf applied, the spring elongates one inch. If local acceleration of gravity is 32.2 ft/s2, what distance, in inches, did the spring elongate?

Respuesta :

fabianb4235

Answer:

x=2.19in

Explanation:

This is the equation that relates the force and displacement of a spring

F=Kx

m=mass=12.5lbx1slug/32.14lb=0.39slug

F=mg=0.39*32.2=12.52Lbf

then we calculate the spring count in lbf / ft

K=F/x

K=5.7lbf/1in=5.7lbf/in=68.4lbf/ft

Finally we calculate the displacement with the initial equation

X=F/k

x=12.52/68.4=0.18ft=2.19in

lhmarianateixeira

In this exercise we want to calculate how much the spring was elogate, for this we have to be:

the spring stayed x=2.19in elogante

organizing the information given in the statement we have that:

  • mass of 12.5 lb
  • force each 5.7 lbf applied
  • acceleration of gravity is 32.2 ft/s

Recalling the basic equation of the spring we find that:

[tex]F=Kx[/tex]

Where:

  • F is the applied force
  • K is the spring constant
  • X is how much the spring has been elongated

So calculating the force we have:

[tex]F=mg\\=0.39*32.2\\=12.52[/tex]

Putting the value of the force in the given formula:

[tex]K=F/x\\K=5.7/1\\=5.7\\X=F/k=12.52/68.4\\=0.18ft\\=2.19in[/tex]

See more about spring at brainly.com/question/4433395

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