As per the question,
Stiffness of the spring(k)=150kN/m=150000N/m(k)=150 kN/m=150000 N/m(k)=150kN/m=150000N/m
Let the mass of the car = m,
Let the acceleration be=a
Extension in the spring =x=x=x
Now, applying the force balance equation,
⇒ma=kx\Rightarrow ma=kx⇒ma=kx
⇒a=kxm=150kNm\Rightarrow a=\dfrac{kx}{m}=\dfrac{150kN}{m}⇒a=mkx=m150kN
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