Question

A 60 kg driver gets into an empty taptap to start the day's work. the springs compress 1.5Ã10â2 m . what is the effective spring constant of the spring system in the taptap?

Answer 1

The force applied to the spring is the weight of the driver:
$W=mg=(60 kg)(9.81 m/s^2)=588.6 N$
Under this force, the spring compresses by $x=1.5 \cdot 10^{-2}m$. We can find the spring's constant k by using Hook's law:
$F=kx$
And so
$k= \frac{F}{x}= \frac{588.6 N}{1.5 \cdot 10^{-2}m}=39240 N/m$