Question

A block of mass 2 kg is compressed against a spring of spring constant$100\mathrm{N}{\mathrm{m}}^{-1}$such that the compression in the spring is $20\mathrm{cm}$, from here the block is released from rest as shown in the figure.
Determine the distance from A where the block falls $[\mathrm{Take}\mathrm{g}=10\mathrm{m}{\mathrm{s}}^{-2}]$.

Answer 1

Step 1: Given data:

Mass of block, $m=2\text{kg}$

Spring constant, $k=100{\text{Nm}}^{-1}$

Compression in spring, $x=0.2\text{m}$

Assume, $g=9.8{\text{ms}}^{-2}$

Height , $H=10\text{m}$

Let v be the velocity of the block when it leaves the spring.

Step 2: Apply conservation of energy

When it leaves the spring, from the energy conservation,

Kinetic energy = Elastic potential energy
$\frac{1}{2}m{v}^2=\frac{1}{2}{\mathrm{kx}}^2$

From the above equation,

$\begin{array}{rcl}v& =& \sqrt{\frac{k}{m}}x\\ & =& \sqrt{\frac{100}{2}}\times 0.2\\ & =& 1.414{\text{ms}}^{-1}\end{array}$

Step 3: Concept of projectile:

If an object is projected from a height $H$ with a velocity $v$, then the maximum range of the projectile is given by the expression

$R_{max}=\frac{v}{g}\sqrt{v^2+2gH}$; g=acceleration due to gravity.

Substitute the known values,

$\begin{array}{rcl}{R}_{max}& =& \frac{1.414}{9.8}\sqrt{1.{414}^2+(2}\times 9.8\times 10)\\ & =& 2.03\text{m}\\ & \approx & 2\text{m}\end{array}$

Therefore, the block will land 2m away from point A.