A weightless spring of force constant 2.50 N/cm is 15.0 cm long when nothing is attached to it. It is now used to pull horizontally on a 12.5 kg box on a smooth horizontal floor. You observe that the box starts from rest and moves 96.0 cm during the first 1.60 s of its motion. How long is the spring during this motion?

The length of the spring during the motion is 16.875 cmHow do I determine the length of the spring?We'll begin by obtaining the velocity during the first 1.60 s. This is illustrated as follow:Distance = 96 cm = 96 / 100 = 0.96 mTime = 1.60 sVelocity =?Velocity = distance / timeVelocity = 0.96 / 1.6Velocity = 0.6 m/sNext, we shall determine the acceleration.Initial velocity (u) = 0 m/sFinal velocity (v) = 0.6 m/sTime (t) = 1.6 sAcceleration (a) =? a = (v – u) / ta = (0.6 – 0) / 1.6a = 0.6 / 1.6a = 0.375 m/s²Next, we shall determine the force.Mass (m) = 12.5 kgAcceleration (a) = 0.375 m/s²Force (F) =?Force = mass × acceleration Force = 12.5 × 0.375F = 4.6875 NNext, we shall determine the extension of the spring.Force (F) = 4.6875 NSpring constant (K) = 2.50 N/cmExtension (e) =?F = KeDivide both sides by Ke = F / Ke = 4.6875 / 2.5e = 1.875 cmFinally, we shall dtermin the length of the spring. This can be obtained as follow:Original length = 15 cmExtended length = 1.875 cmLength of spring =?Length of spring = 15 + 1.875 Length of spring = 16.875 cmLearn more about length of spring:https://brainly.com/question/12982193https://brainly.com/question/17029566https://brainly.com/question/12643730#SPJ1