A small block of mass 100 g is pressed again a horizontal spring fixed at one end to compress the sprign through 5.0 cm. The spring constant is 100 N/m. When released, the block moves hroizontally till it leaves the spring. Where will it hit the ground 2 m below the spring?

Explanation

When the block is released, it moves horizontally with speed V till it leaves the spring.
By energy conservation, `1/2kx^2=1/2mV^2`
`V^2=(kx^2)/(m)impliesV=sqrt((kx^2)/(m))`
Time of flight, `t=sqrt((2H)/(g))`
So horizontal distance travelled from the free end of the spring is
`Vxxt=sqrt((kx^2)/(m))xxsqrt((2H)/(g)a)`
`=sqrt((100xx(0.05)^2)/(0.1))xxsqrt((2xx2)/(10))=1m`
So at a horizontal distance of `1m` from the free end of the spring.