Question

a 22-kg chimpanzee hangs from the end of a horizontal broken branch the branch sags downward through a vertical distance of 13 cm treating the branch as a spring satisfying hooke's law what is its spring constant

Answer 1

The spring constant of the branch satisfying Hooke's law is 1,658.46 N/m.

The given parameters;

• mass of the chimpanzee, m = 22 kg
• extension of the branch, x = 13 cm = 0.13 m

The spring constant of the branch satisfying Hooke's law is calculated as follows;

F = kx

mg = kx

$k = \frac{mg}{x} \\\\k = \frac{22 \times 9.8}{0.13} \\\\k = 1658.46 \ N/m$

Thus, the spring constant of the branch is 1,658.46 N/m.

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