Final answer:
The trampoline spring with a stretch of 0.25m and a spring constant of 250N/m stores 7.8125 Joules of elastic potential energy.
Step-by-step explanation:
The question asks how much elastic potential energy is stored in a trampoline spring when it is stretched by 0.25m and has a spring constant of 250N/m. In physics, the formula for the elastic potential energy stored in a spring is U = ½kx², where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
For this particular trampoline spring, with k = 250 N/m and x = 0.25 m, we can calculate the elastic potential energy as:
U = ½ * (250 N/m) * (0.25 m)²
= ½ * 250 * 0.0625
= ½ * 15.625
= 7.8125 Joules
Therefore, 7.8125 Joules of elastic potential energy is stored in the spring.