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kingkos
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.
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
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
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