A 5-kg block rests on a smooth inclined plane of inclination `37^(@)` as shown in the diagram if the extension of spring is 6cm, find (a) the spring constant (b). If the block is pulled dow the incline 4 cm from its equilibrium position and released, find the initial acceleration of the block.

Explanation

(a). In equilibrium
`x_(0)=6cm=0.06cm`
`5gsin37^(@)=kx_(0)`
`5xx10xx(3)/(5)=kxx0.06`
`k=(30)/(0.06)=500N//m`
(b).
`k(x_(0)+x)-5gsin37^(@)=ma=5a`
`500((6+4))/(100)-30=5a`
`a=4m//s^(2)`