A block of mass 1 kg is suspended from a spring of constant 100 N/m. initially the spring is unstretched. The block is then released. What is the acceleration (in `m//s^(2)`) of the block when its velocity is 0.6 m/s in the downwards direction ?

Answer

8

Explanation

`mgx-1/2 kx^(2)=1/2 mv^(2)`
`1xx10x-50 x^(2)=1/2 xx1xx(0.6)^(2) `

`50x^(2) -10x+18/100=0`
`x=(10pmsqrt(100-4xx50xx9/50))/100=18/100,2/100`
mg-kx=ma
`a=1xx10-100xx18/100=-8 m//s^(2)`