Question

A 69 kg driver gets into an empty taptap to start the day's work. The springs compress 2×10−2 m . What is the effective spring constant of the spring system in the taptap? Enter the spring constant numerically in newtons per meter using two significant figures.

Answer 1

Answer:

$3.4\cdot 10^4 N/m$

Explanation:

The spring system in the taptap obey's Hooke's law, which states that:

$F=kx$

where

F is the magnitude of the force applied

k is the spring constant

x is the compression/stretching of the spring

In this problem:

- The force applied is the weight of the driver of mass m = 69 kg, so

$F=mg=(69 kg)(9.8 m/s^2)=676.2 N$

- The compression of the spring is

$x=2\cdot 10^{-2} m=0.02 m$

So, the spring constant is

$k=\frac{F}{x}=\frac{676.2 N}{0.02 m}=3.4\cdot 10^4 N/m$