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sweet-ann
17 days ago
11

"a very light ideal spring stretches by 21.0 cm when it is used to hang a 135-n object. what is the weight of a piece of electro

nic equipment that would stretch the spring 44.9 cm by if you hung the equipment using the spring?
Physics
2 answers:
Keith_Richards [2.2K]17 days ago
6 0
In the first case, once equilibrium is established,
mg = kx
135 = k(0.21) [Converted to SI units]
k = 135/(0.21)

In the second case, similarly when reaching equilibrium,
mg = kx
mg = [(135)(.449)]/(0.21)
mg = weight of the electronic equipment = approximately 289 N.
Keith_Richards [2.2K]17 days ago
3 0

Answer:

F= 288.7 N

Explanation:

According to Hooke's Law

"the strain or deformation experienced by an elastic body is directly proportional to the stress applied to it."

We can calculate the spring constant using Hooke's Law:

F = k*x

therefore,

k = F/x

k = 135N/0.21m

k = 643 N/m

When the fish stretches the spring by 0.449m, its weight would be

F =k*x

F= 643N/m * 0.449m

F= 288.707 N

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