Assuming given quantities are correct or taken as below.
A. We know from Hooke's Law that equation for restoring Force #F# acting on the spring is
#F=−kx# .....(1)
where #k# is Spring Constant, #x# is deformation.
Inserting given values and converting all in SI units we get
#F=−1.2xx10^4xx6/100#
#F=−7.2xx10^2N#
(#-ve)# sign shows that force is restoring force or is opposite to the direction of deformation.
B. Elastic potential energy is given as #PE=1/2kx^2# .....(2)
At equilibrium position #x=0#, therefore #PE_0=0#
#PE_6# in the stretched position#=1/2xx1.2xx10^4xx(6/100)^2#
#=21.6J#
Now change in #PE="Final" -"Initial"# ......(3)
#PE_6-PE_0=21.6 -0=21.6J#
C. #PE_3# in the stretched position#=1/2xx1.2xx10^4xx(3/100)^2#
#=5.4J#
Using equation (3)
#PE_3-PE_6=5.4-21.6=-16.2J#
D. Using equation (3) and calculated #PE# in C. above
#PE_0-PE_3=0-5.4=-5.4J#
#-ve" "PE# shows that Elastic potential energy is decreasing and getting converted into kinetic energy of the spring.
