Given : kA=300 N/m,kB=400 N/m Let when the combination of springs is compressed by force F. Spring A is compressed by x. Therefore compression in spring B xB=(8.75−x)cmF=300×x=400(8.75−x) Solving we get, x=5 cm xB=8.75−5=3.75 cm EBEA=21kB(xB)221kA(xA)2=400×(3.75)2300×(5)2=34