A box of mass m = 1.80 kg is dropped from rest onto a massless, vertical spring with spring constant k = 2.00 ✕ 102 N/m that is initially at its natural length. How far is the spring compressed by the box if the initial height of the box is 2.25 m above the top of the spring?

Question

Drake Hammond

Physics

6 February, 04:56

A box of mass m = 1.80 kg is dropped from rest onto a massless, vertical spring with spring constant k = 2.00 ✕ 102 N/m that is initially at its natural length. How far is the spring compressed by the box if the initial height of the box is 2.25 m above the top of the spring?

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Answers (1)

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Waldo · Answer 1

spring compressed is 0.724 m

Explanation:

given data

mass = 1.80 kg

spring constant k = 2 * 10² N/m

initial height = 2.25 m

solution

we know from conservation of energy is

mg (h+x) = 0.5 * k * x² ... 1

here x is compression in spring

so put here value in equation 1 we get

1.8 * 9.8 * (2.25+x) = 0.5 * 2 * 10² * x²

solve it we get

x = 0.724344

so spring compressed is 0.724 m