A vertically mounted spring (k = 750 N/m) is compressed by 35.0 cm relative to its unstrained length. A mass (m = 0.36 kg) is placed at rest against the spring. When the spring is released, the mass is launched vertically in the air. How high from release point the mass can reach?

When the spring is released, the mass is launched vertically in the air. We need to find the height attained by the mass at this position. On applying the conservation of energy as :

Energy stored in the spring = change on potential energy

[tex]\dfrac{1}{2}kx^2=mgh[/tex]

[tex]h=\dfrac{kx^2}{2mg}[/tex]

[tex]h=\dfrac{750\ N/m\times (0.35\ m)^2}{2\times 0.36\ kg\times 9.8\ m/s^2}[/tex]

h = 13.02 meters

So, the mass will reach a height of 13.02 meters. Hence, this is the required solution.

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-10-29T19:38:44+02:00

The maximum height reached by the mass when the spring is released is 13.02 m.

The given parameters;

spring constant, k = 750 N/m
extension of the spring, x = 35 cm = 0.35 m
mass attached, m = 0.36 kg

Apply the principle of conservation energy to determine the maximum height reached by the mass when the spring is released.

mgh = ¹/₂kx²

where;

h is the maximum height reached by the mass
g is acceleration due to gravity

[tex]h = \frac{kx^2}{2mg} \\\\h = \frac{(750) \times (0.35)^2}{2(0.36\times 9.8)} \\\\h = 13.02 \ m[/tex]

Thus, the maximum height reached by the mass when the spring is released is 13.02 m.

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