1 kg block collides with a horizontal massless spring of force constant `2 N//m`. The block compresses the spring by 4m. If the coefficient of kinetic friction between the block and the· surface is 0.25, what was the speed of the block at the instant of collision? (take `g = 10 m//s^(2)`)

Explanation

When the block compresses the spring, let x be the amount of compression, i.e x = 4m.
Let v = velocity oof the block when it collides with the spring.
Loss in KE of the block = (gain in elastic potential energy of the spring) + (work done against friction)
`Rightarrow 1/2 mv^(2) - 0 = 1/2 kx^(2) + mu mgx = 1/2 xx 1 xx v^(2)`
` = 1/2 (2) (4)^(@)+ 0.25 xx 1 xx9.8 xx 4`
`v^(2) = 516 Rightarrow v = sqrt(516) = 7.18 m/s`.