How Far Was the Spring Compressed in the Physics Problem?

AI Thread Summary
A 0.75-kg object compresses a spring with a spring constant of 50 N/m and is released, achieving a speed of 2.30 m/s. The relationship between kinetic energy and spring force is used to determine the compression distance. The calculation shows that the spring was compressed by approximately 3.97 cm. There is a discussion on whether to equate force to kinetic energy or to relate potential energy stored in the spring to the object's kinetic energy. The final answer confirms the compression distance as correct.
wondermoose
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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!
 
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hi wondermoose! :smile:
wondermoose said:
s= -(1/2mv^2)/k

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

s= -0.0397 m

or 3.97cm

looks good! :biggrin:
 
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!
 
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?
 
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