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

• A. $0.4m{s}^{-1}$
• B. $4m{s}^{-1}$
• C. $0.8m{s}^{-1}$
• D. $8m{s}^{-1}$

Answer 1

The correct option is D $8m{s}^{-1}$

Let the velocity of the block at the time of collision is $v$.
The kinetic energy of the block at the instant of collision is given by applying work energy theorem from point of collision to rest position

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

$\frac{1}{2}{v}^2=\frac{1}{2}\left(2.75\right)4^2+\left(0.25\right)\left(10\right)\left(4\right)$
$v^2=54+10=64$
$v=8m{s}^{-1}$