# Mass on a spring.

A 9.3 kg block is dropped onto a spring of spring constant 1086 N/m from a height of 800 cm. When the block is momentarily at rest, the spring has been compressed by 50 cm. Find the speed of the block when the compression of the spring is 10 cm. The acceleration of gravity is 9.81 m/s^2. Answer in units of m/s.

Equations...
1/2kx^2
k = 1/2mv^2
u = mgh

my problem is I dont understand which equations to use in which situations,..

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This is simply a transformation of energy question.

Right at the start before the block is dropped, only gravitation potential energy (Eg) is present. Then all of the Eg is turned into kinetic energy (Ek) as the block falls. Just before the block hits the spring, all of the Eg is turned into Ek. Then the block impacts the spring and compressing it, turning Ek into Elastic potential energy (Ee).

Therefore, it would:

Eg-->Ek-->Ee

Using this transformation method, it would be very simple to solve this.

If I wasn't very familiar with the whole transformation of energy...how would I do this?

I tried 1/2mv^2 = 1/2kx^2 and it comes out wrong..