A vertical ideal spring is mounted on the floor and has a springconstant of 162

Question

A vertical ideal spring is mounted on the floor and has a springconstant of 162 N/m. A 0.80-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.

Explanation / Answer

(a)    as the block rests stationary on the spring,the downward-directed weight balances the upward-directed restoringforce
   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 = k x    x = m g / k       = ........ m (b)    the conservation of mechanical energy statesthat the final total mechanical energy Ef is equalto the initial total mechanical
   energy Eo    the expression for the total mechanicalenergy for an object on a spring is given by
   E = (1 / 2) m v2 + (1 / 2) I2 + m g h + (1 / 2) k x2    so that we have    (1 / 2) m vf2 + (1 /2) I f2 + m g hf + (1 / 2) kxf2 = (1 / 2) m vo2 +(1 / 2) I o2 + m g ho + (1 /2) k xo2    the block does not rotate, so the angularspeeds wf and w0 are zero. Sincethe block comes to a momentary 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 g hf + (1 / 2) kxf2 = m g ho    (1 / 2) kxf2 = m g (ho-hf)      the term (ho-hf) is the amount by which the spring hascompressed,    (ho -hf) =xf    making this substitution into the simplifiedenergy-conservation equation gives    (1 / 2) k xf2 = mg (ho -hf)                      = m g xf    (1 / 2) k xf = m g    xf = 2 m g / k        = ........ m    m g hf + (1 / 2) kxf2 = m g ho    (1 / 2) kxf2 = m g (ho-hf)      the term (ho-hf) is the amount by which the spring hascompressed,    (ho -hf) =xf    making this substitution into the simplifiedenergy-conservation equation gives    (1 / 2) k xf2 = mg (ho -hf)                      = m g xf    (1 / 2) k xf = m g    xf = 2 m g / k        = ........ m    (1 / 2) kxf2 = m g (ho-hf)      the term (ho-hf) is the amount by which the spring hascompressed,    (ho -hf) =xf    making this substitution into the simplifiedenergy-conservation equation gives    (1 / 2) k xf2 = mg (ho -hf)                      = m g xf    (1 / 2) k xf = m g    xf = 2 m g / k        = ........ m

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