Question

An ejector mechanism consists of a helical compression spring having a spring constant of $K=981\times {10}^3\frac{N}{m}$. It is pre-compressed by 100 mm from its free state. If it is used to eject a mass of 100 kg held on it, the mass will move up through a distance of

• A. 100
• B. 5000 mm
• C. 581 mm
• D. 1000 mm

Answer 1

The correct option is B 5000 mm

Method I:
$\frac{1}{2}K{x}^2$ will be converted into potential energy (taking reference as compressed condition)
$\frac{1}{2}K{x}^2=mgh$
$\frac{1}{2}\times 981\times {10}^3\times (0.1{)}^2=100\times 9.81h$
$h=5m=5000mm$
Method II:
When spring will reach its natural length, K.E. of the block of mass 100 kg held on it.
(taking reference at natural length of spring)
$\frac{1}{2}k{x}^2-mgx=\frac{1}{2}m{V}^2[x=0.1m]$
and this K.E will be converted into mgh.
Total lift of block =h+x
$\frac{1}{2}\times 981\times {10}^3\times .1-100\times 9.81\times .1=mgh=100\times 9.81\times h$
h = 4.9 m
$\Rightarrow h+x=4.9+0.1=5m$