Question

A spring with an unknown force constant is compressed by an object of an unknown mass.(the diagram looks like the spring is vertical and the mass is placed perpendicular to the cross-section of the spring.) The original length of the spring is 60 cm. When the mass was placed above the spring, the length was compressed to 50 cm.(meaning the change in the length is 10 cm). Calculate the spring force constant. Diameter of the spring cross section is 8 cm. Thanks!!

Answer 1

according to Hooke's Law, f=-kx, where f is the force of the spring, k is the spring constant (force constant in your post), and x is THE DIFFERENCE (subtraction) BETWEEN THE 'RELAXED LENGTH' OF THE SPRING (how long it is with no mass attached) AND THE COMPRESSED OR EXTENDED LENGTH WHEN A MASS IS ATTACHED.

when the spring is compressed, it tries to push the mass back to the relaxed length, and when it is extended it tries to pull the mass back to relaxed length.

your explanation says that the spring is being compressed in the vertical direction by a distance of 10 centimeters from relaxed length. that means the mass is pulled down by gravity to compress the spring. ebcause the mass is not moving, there is NO NET FORCE on the mass (f=ma=m*0=0). this means that the force of gravity pulling down on the mass is equal to the force the spring exerts upward on the mass--the two forces are of equal magnitude but opposite direction (meaning opposite signs).

we can use the equation above to solve for k IN TERMS OF M, but without knowing m, we cannot know the force exerted on the mass by gravity and therefore cannot find a numerical value for k without more information. with k or m given, we can solve for the other, or if the spring is in oscillating motion, then we can use the period oscillation to get more information.

Answer 2

u need to have either of the mass or the spring constant to get this
the area of cross section is useless