Question

A vertically mounted spring (k = 850 N/m) is compressed by 31.0 cm relative to its unstrained length. A mass (m = 0.24 kg) is placed at rest against the spring. When the spring is released, the mass is launched vertically in the air. How high from the release point can the mass reach? g

Answer 1

Answer:

h = 1.74m

Explanation:

Given that k = 850N/m, x = 31.0cm = 0.31m, m = 2.4kg.

The act of stretching the spring stores potential energy in the spring. On attaching a mass of 0.24kg to the spring in its stretched state and releasing it, the stored potential energy in the spring is converted into the gravitational potential of the spring.

So

1/2kx² = mgh

h = ? g = 9.8m/s²

h = 1/2×kx²×1/mg

h = 1/2×850×0.31²×1/(2.4×9.8)

h = 1.74m

Answer 2

The maximum height reached by the mass when the spring is released is 17.37 m.

The given parameters;

• spring constant, k = 850 N/m
• extension of the spring, x = 31.0 cm = 0.31 m
• mass of the spring, m = 0.24 kg

The height attained by the mass is calculated  by applying the principle of conservation of energy;

mgh = ¹/₂kx²

$h = \frac{kx^2}{2mg} \\\\h = \frac{(850) \times (0.31)^2}{2\times 0.24 \times 9.8} \\\\h = 17.37 \ m$

Thus, the maximum height reached by the mass when the spring is released is 17.37 m.

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