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3 years ago

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A block of mass 0.249 kg is placed on top of a light, vertical spring of force constant 4 975 N/m and pushed downward so that th

e spring is compressed by 0.107 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? (Round your answer to two decimal places.)

1 answer:

Brut [27]3 years ago

7 0

Answer:

h = 11.67 m

Explanation:

Mass, m = 0.249 kg

Spring constant, k = 4975 N/m

Compression, x = 0.107 m

Energy stored in the spring,W, is:

W = 0.5kx²

W = 0.5*4975*0.107²

W = 28.48N·m

The potential energy is given by the formula:

PE = mgh

PE = 0.249*9.8*h

PE = 2.4402h

According to energy conservation principle, Potential Energy = spring energy.

That is, PE = W

2.4402h = 28.48

h = 28.48/2.4402

h = 11.67 m

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