A vertical ideal spring is mounted on the floor and has a springconstant of 188

Question

A vertical ideal spring is mounted on the floor and has a springconstant of 188 N/m. A 1.00-kg block is placed on the spring in twodifferent ways. (a) In one case, the block isplaced on the spring and not released until it rests stationary onthe spring in its equilibrium position. Determine the amount(magnitude only) by which the spring is compressed.(b) In a second situation, the block is releasedfrom rest immediately after being placed on the spring and fallsdownward until it comes to a momentary halt. Determine the amount(magnitude only) by which the spring is now compressed.

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Explanation / Answer

(a)    as the blockrests stationary on the spring, the downward-directed weightbalances the upward-directed restoring force
   from the spring
   the magnitude of the weight is mg, and themagnitude of the restoring force is given
   F = k x
   without the minus sign as kx
   thus, we have    m g = kx    x = m g /k       =........ m (b)    theconservation of mechanical energy states that the final totalmechanical energy Ef is equal to the initial totalmechanical
   energy Eo    the expressionfor the total mechanical energy for an object on a spring is givenby
   E = (1 / 2) m v2 + (1 / 2) I2 + m g h + (1 / 2) k x2    so that wehave    (1 / 2) mvf2 + (1 / 2) If2 + m g hf + (1 / 2) kxf2 = (1 / 2) m vo2 +(1 / 2) I o2 + m g ho + (1/ 2) k xo2    the block doesnot rotate, so the angular speeds wf andw0 are zero. Since the block comes to amomentary halt on the
   spring and is released from rest, thetranslational speeds vf and v0 are alsozero
   because the spring is initially unstrained, theinitial displacement x0 of the spring is likewisezero
   thus, the above expression can be simplified asfollows    m ghf + (1 / 2) k xf2 = m gho    (1 / 2) kxf2 = m g (ho-hf)      the term(ho -hf) is the amount by which thespring has compressed,    (ho -hf) =xf    makingthis substitution into the simplified energy-conservation equationgives    (1 / 2) kxf2 = m g (ho-hf)                      = m g xf    (1 / 2) kxf = m g    xf = 2 m g /k        = ........m
   m ghf + (1 / 2) k xf2 = m gho    (1 / 2) kxf2 = m g (ho-hf)      the term(ho -hf) is the amount by which thespring has compressed,    (ho -hf) =xf    makingthis substitution into the simplified energy-conservation equationgives    (1 / 2) kxf2 = m g (ho-hf)                      = m g xf    (1 / 2) kxf = m g    xf = 2 m g /k        = ........m
   (1 / 2) kxf2 = m g (ho-hf)      the term(ho -hf) is the amount by which thespring has compressed,    (ho -hf) =xf    makingthis substitution into the simplified energy-conservation equationgives    (1 / 2) kxf2 = m g (ho-hf)                      = m g xf    (1 / 2) kxf = m g    xf = 2 m g /k        = ........m

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