How Far Was the Spring Compressed in the Physics Problem?

[wondermoose]

Homework Statement

A 0.75-kg object rests on a horizontal frictionless surface. It is in a position such that it is compressing a spring with a spring constant of 50 N/m. If the object is released, the object leaves the spring at a speed of 2.30 m/s. How far was the spring compressed?

Homework Equations

K=1/2mv^2
F(spring)=-ks
U(spring)=1/2k(s^2)

The Attempt at a Solution

Okay so I attempted to find the distance of compression by relating the K=1/2mv^2 to the force of a spring (-ks)

1/2mv^2=-ks

(1/2mv^2)/k=-s

s= -(1/2mv^2)/k

s= -((1/2(0.75)(2.3)^2))/50

s= -0.0397 m

or 3.97cm

Thanks!

 

[tiny-tim]

hi wondermoose!

s= -(1/2mv^2)/k

s= -((1/2(0.75)(2.3)^2))/50

s= -0.0397 m

or 3.97cm

looks good!

 

[wondermoose]

Wow, really? That was a slightly modified version of the attempt I made on my test... I didn't actually think I was that close. Thanks!

 

[divineyang]

is it correct to equate the force to the kinetic energy of the object? shouldn't you relate the potential energy stored in the spring to the kinetic energy of the object instead?