Question

You, a 70 kg person, leap from a 10 m tall building and land feet first on a trampoline. The center of the trampoline where you land sinks a total of 0.5 m before springing you back into the air. What is the spring constant for the trampoline?

Answer 1

Answer:

1373.4 N/m

Explanation:

Hooke's law states that the extension of a spring and force are related by the expression, F=kx where k is spring constant, x is extension of spring and F is the applied force. Making k the subject of the formula then

$k=\frac {F}{x}$

Also, F=gm hence the above formula is modified as

$k=\frac {gm}{x}$

Taking g as 9.81 m/s2 , x as 0.5 m and m as 70 kg then

$k=\frac {9.81\times 70 kg}{0.5m}=1373.4 N/m$