Question

A $2\text{kg}$ block slides on a horizontal floor with a speed of $4\text{m/s}$. It strikes an uncompressed spring and compresses it till the block is motionless. The kinetic friction force is $15\text{N}$ and spring constant is $10,000\text{N/m}$. The spring compresses by

• A. $5.5\text{cm}$
• B. $2.5\text{cm}$
• C. $11.0\text{cm}$
• D. $8.5\text{cm}$

Answer 1

The correct option is A $5.5\text{cm}$

The total kinetic energy possessed by the block goes into the spring elastic energy and the work done against friction.
Let $x$ be the compression of the spring,
$v$ be the initial velocity of the block and $f_r$ be the kinetic friction force

Applying work-energy theorem :
$W_{net}=\Delta KE$
$W_{friction}+W_{spring}=\Delta KE$
$\Rightarrow -f_{r}x-\frac{1}{2}k{x}^2=-\frac{1}{2}m{v}^2$
(Final velocity of block = 0, initial compression of spring = 0 )
$\Rightarrow 5000x^2+15x-16=0$

Solving the equation, we get
$x=0.055\text{m}=5.5\text{cm}$