Question

1. a toy airplane ejects its 8.4 g pilot vertically using a spring with a spring constant of 5.2 x 10² n/m. if the spring is initially compressed by 5.2 cm down,

a) calculate the speed of the pilot as it ejects upward from the airplane.
b) determine the maximum height that the pilot will reach.

Explanation:

Step1: Convert units

Convert mass \(m = 8.4g=8.4\times10^{- 3}kg\), spring compression \(x = 5.2cm = 0.052m\)

Step2: Use conservation of mechanical energy for part (a)

The elastic potential energy of the spring \(E_{p}=\frac{1}{2}kx^{2}\) is converted into kinetic energy \(E_{k}=\frac{1}{2}mv^{2}\) at the ejection point (assuming no non - conservative forces).

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Substitute \(k = 5.2\times10^{2}N/m\), \(m = 8.4\times10^{-3}kg\) and \(x = 0.052m\)

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Step3: Use kinematic equation for part (b)

At the maximum height, the final velocity \(v_f = 0\). Using the kinematic equation \(v_f^{2}-v^{2}=- 2gh\) (where \(a=-g=-9.8m/s^{2}\))

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Substitute \(v\approx12.94m/s\) and \(g = 9.8m/s^{2}\)

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Answer:

a) The speed of the pilot as it ejects upward from the airplane is approximately \(v = 12.9m/s\)
b) The maximum height that the pilot will reach is approximately \(h = 8.54m\)