An ideal spring of force constant k is hung vertically from the ceiling, and a held object of mass m is attached to the loose end. You carefully and slowly ease that mass down to its equilibrium position by keeping your hand under it until it reaches that position. (a) Show that the spring's change in length is given by d=mg/k. (b) Show that the work done by the spring is Wsp=-m^2g^2/2k. (c) Show that the work done by gravity is Wg=m^2g^2/k. Explain why these two works do not add to zero. (d) Show that the work done by your hand is Whand=-m^2g^2/2k and that the hand exerted an average force of half the object's weight.

Question

An ideal spring of force constant k is hung vertically from the ceiling, and a held object of mass m is attached to the loose end.  You carefully and slowly ease that mass down to its equilibrium position by keeping your hand under it until it reaches that position. (a) Show that the spring's change in length is given by d=mg/k.  (b)  Show that the work done by the spring is Wsp=-m^2g^2/2k. (c) Show that the work done by gravity is Wg=m^2g^2/k.  Explain why these two works do not add to zero.  (d) Show that the work done by your hand is Whand=-m^2g^2/2k and that the hand exerted an average force of half the object's weight.

anonymous · Answer

a)  F = - k d = - m g 
so d = (mg/k)

b) work = integral (k x) dx = (1/2) k x^2 = (1/2) k d^2 = (1/2) k (mg/k)^2
 = (mg)^2 / 2k 

c) work = F d is constant, so = mgd

d) subtract (b) from (c) and find the difference, to bring it down "gently," the hand split the work with the spring.  Neat.