a force of 750n streches a certain spring a distance of 0.150 m.what is the pe of the spring when a 60kg mass hangs vertically from it.

Question

anonymous · Answer

m= 60 kg
F=750 N
X=.150 m

P.e=?

as P.e=1/2 k$$x ^{2}$$

but k=?

for k use 

F=-kx

K=-F/x=750/.150=-5000

put it in P.e equation i.e

P.E=1/2k$$x ^{2}$$=0.5 x (-5000) x $$0.150^{2}$$=-56.25 joules

anonymous · Answer

Correct until final part. The x is not .15m with the 60kg mass. 

x=F/-k
  =mg/k
  =60g/5000
  ==.11772

Therefore P.E.= .5 * 5000 * .11772^2=34.645J

Note minus sign just to show force in opp. dir. to displacement

anonymous · Answer

k comes from ?

anonymous · Answer

As you showed, F=-kx (Hooke's Law). The first part is in order to get k, which doesnt change for the second part as  its the same spring