physics
posted by Anonymous on .
In 0.511 s, a 13.1kg block is pulled through a distance of 3.13 m on a frictionless horizontal surface, starting from rest. The block has a constant acceleration and is pulled by means of a horizontal spring that is attached to the block. The spring constant of the spring is 466 N/m. By how much does the spring stretch?

F=ma=kx
k=ma/x
so what is acceleration a?
vaverage=3.13m/.511s
vfinal then must be twice that, or 6.26/.511
but we know
Vfinal^2=Vi^2+2ad so solve for a
Then, solve for k from k=ma/x