Question

A $16kg$ block moving on a frictionless horizontal surface with a velocity of $4m{s}^{-1}$compresses an ideal spring and comes to rest. If the force constant of the spring be $100N{m}^{-1}$, then the spring is compressed by

• A. $1.6m$
• B. $4m$
• C. $6.1m$
• D. $3.2m$

Answer 1

The correct option is A

$1.6m$

Step 1: Given data

Mass of the block, $m=16kg$

Velocity of the block, $v=4m/s$

Spring constant, $k=100N/m$

Step 2. Find the compression in spring, $\mathit{x}$

Kinetic energy is half of the product of the mass and the square of the velocity.

Potential energy of the spring is equal to half of the product of the spring constant and the square of the compression in the spring.

Applying conservation of energy, kinetic energy, $K.E.$ is equal to the potential energy of the spring, $P.E.$

$K.E.=P.E.$

$\Rightarrow$ $\begin{array}{rcl}\frac{1}{2}\left(m{v}^2\right)& =& \frac{1}{2}\left(k{x}^2\right)\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{\left(16\times 4\times 4\right)}{2}& =& \frac{\left(100\times x^2\right)}{2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}100\times x^2& =& 16\times 4\times 4\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{x}^2& =& \frac{\left(16\times 4\times 4\right)}{100}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}x& =& 1.6m\end{array}$

Hence, option (A) is correct.