Question

A 61 kg driver gets into an empty taptap to start the day's work. The springs compress 1.7×10−2 m . What is the effective spring constant of the spring system in the taptap?

Answer 1

Answer:

Spring constant, $k=3.51\times 10^4\ N/m$

Explanation:

Given that,

Mass of the driver, m = 61 kg

Compression in the spring, $x=1.7\times 10^{-2}\ m$

We need to find the effective spring constant of the spring. It can be calculated by equating force of gravity and the force in case of spring such that :

$mg=kx$

k is the spring constant of the spring system

$k=\dfrac{mg}{x}\\\\k=\dfrac{61\times9.8}{1.7\times10^{-2}}\\\\k=3.51\times 10^4\ N/m$

So, the effective spring constant of the spring system in the taptap is $3.51\times 10^4\ N/m$. Hence, this is the required solution.