Final answer:
The potential energy stored in an ideal spring when it is stretched by 1 cm is 0.0003 J.
Step-by-step explanation:
The potential energy stored in an ideal spring when it is stretched by 1 cm can be calculated using the equation for potential energy in a spring: PE = ½kx², where k is the spring constant and x is the displacement. In this case, the spring constant is given as 60 N/m and the displacement is 1 cm, which is equal to 0.01 m. Plugging these values into the equation, we get:
PE = ½(60 N/m)(0.01 m)² = ½(60 N/m)(0.0001 m²) = 0.0003 J
Therefore, the spring stores 0.0003 joules of energy when it is stretched by 1 cm.