Question

a spring attached to a mass is at rest in the initial position (not shown). the spring is compressed in position a and is then released, as shown in position b. which equation describes conservation of energy in position a? e = 1/2 mv² e = mgh e = 1/2 kx² e = 1/2 k2x²

Explanation:

Step1: Identify energy in position A

In position A, the spring - mass system has only elastic potential energy as the mass is at rest (kinetic energy $K = 0$) and there is no change in height (gravitational potential energy $U_g=0$).

Step2: Recall elastic - potential energy formula

The formula for elastic potential energy of a spring is $E_p=\frac{1}{2}kx^{2}$, where $k$ is the spring constant and $x$ is the displacement from the equilibrium position. Here, in position A, the total energy $E$ of the system is equal to the elastic potential energy of the compressed spring. So $E = \frac{1}{2}kx^{2}$.

Answer:

C. $E=\frac{1}{2}kx^{2}$