Question

One end of a spring of force constant $k_{1}$ is attached to the ceiling of an elevator. A block of mass $1.5\, kg$ is attached to the other end. Another spring of force constant $k_{2}$ is attached to the bottom of the mass and to the floor of the elevator as shown in the figure. At equilibrium, the deformation in both the springs is equal and is $40\, cm$. If the elevator moves with constant acceleration upward, the additional deformation in both the springs is $8\, cm$. Find the elevator's acceleration $\left(g=10\, m s ^{-2}\right)$.

Answer 1

At equilibrium, $\left(k_{1}+k_{2}\right) x=m g$
$\Rightarrow k_{1}+k_{2}=\frac{m g}{x}$ ...(i)
$\left(x=40\, cm , x_{1}=48\, cm \right)$
In case of constant acceleration,
$\left(k_{1}+k_{2}\right) x_{1}-m g=m a$ ...(ii)
From Eqs. (i) and (ii), we get $a=2\, ms ^{-2}$