When a spring is stretched by 2 cm, the energy stored in it is proportional to the square of the displacement (Hooke's law for elastic potential energy). If the spring stores 100 J of energy at 2 cm stretch, stretching it further by another 2 cm (total 4 cm) increases the energy stored to 400 J. The increase in energy due to this additional stretch of 2 cm is 300 J. This is derived from the formula for elastic potential energy stored in a spring:
U=12kx2U=\frac{1}{2}kx^2U=21kx2
where UUU is the potential energy, kkk is the spring constant, and xxx is the displacement from the spring's natural length. For the initial 2 cm stretch (x=2x=2x=2 cm),
U1=12k(2)2=100 JU_1=\frac{1}{2}k(2)^2=100,\text{J}U1=21k(2)2=100J
For the total 4 cm stretch,
U2=12k(4)2=400 JU_2=\frac{1}{2}k(4)^2=400,\text{J}U2=21k(4)2=400J
The increase in energy stored is:
ΔU=U2−U1=400−100=300 J\Delta U=U_2-U_1=400-100=300,\text{J}ΔU=U2−U1=400−100=300J
Thus, stretching the spring by an additional 2 cm beyond the initial 2 cm increases the stored energy by 300 J.
