Question

A $1.0kg$ block collides with a horizontal weightless spring of force constant $2.75N{m}^{-1}$ as shown in figure. The block compresses the spring $4.0m$ from the rest position. If the coefficient of kinetic friction between the block and horizontal surface is $0.25$ , the speed of the block at the instant of collision is :

Answer 1

Let the velocity of the block at the time of collision is $v$.

The kinetic energy of the block is given as,

$\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2+{\mu }_{k}mgx$

$\frac{1}{2}\times 1\times v^2=\frac{1}{2}\times 2.75\times {\left(4\right)}^2+\left(0.25\right)\times 1\times 9.8\times 4$

$v^2=63.6$

$v=7.975m/s$

Thus, the speed of block at the instant of collision is $7.975m/s$.