Question

A 0.10 meter spring is stretched from equilibrium to position A at 0.15 meters and then to Position B at 0.20 meters. Compared with the springs potential energy at A, what is the potential energy at B? Is it:
a) the same
b) twice as great
c) half as great
d) four times as great

Answer 1

b) twice as great

The potential energy of a spring is given by the formula: PE = 1/2kx^2, where k is the spring constant and x is the displacement from equilibrium.

At position A, the spring is stretched by 0.15 meters, so the potential energy at A is PE_A = 1/2k(0.15)^2.

At position B, the spring is stretched by 0.20 meters, so the potential energy at B is PE_B = 1/2k(0.20)^2.

To compare the potential energy at A and B, we can take the ratio of PE_B to PE_A:

PE_B / PE_A = (1/2k(0.20)^2) / (1/2k(0.15)^2) = (0.20)^2 / (0.15)^2 = 1.777

Therefore, the potential energy at B is approximately twice as great as the potential energy at A.