Question

A 68 kgkg driver gets into an empty taptap to start the day's work. The springs compress 2.1×10−2 mm . What is the effective spring constant of the spring system in the taptap?

Answer 1

Answer:

K = 3.1 8 × 10⁴ Nm

Explanation:

Given:

m = 68 Kg

x= 2.1 × 10 ⁻² m (in (2.1×10−2 mm) i think m is repeated just like kgkg in the start of question)

g=9.81 m/s²

K= ?

According to Hooke's Law

F = K x  

also F=mg

⇒ mg = Kx

⇒ K = mg/x

K= 68 Kg × 9.81 m/s² / 2.1 × 10 ⁻² m

K = 3.1 8 × 10⁴ Nm