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?

 

[rwisz]

Your approach to solving this problem is correct. By equating the kinetic energy of the object to the potential energy stored in the spring, you can solve for the distance of compression. Your calculations are also correct, resulting in a compression distance of approximately 3.97 cm. Keep up the good work!