The position of a 2.0-kg mass connected to a spring whose constant is 15 N/m is given by $x = A \cos ( \omega t - \delta )$ , where A = 25 cm and $\delta$= $\pi$/2. The total mechanical energy of the mass at time t = 0.30 seconds is
A) 0 J.
B) 0.34 J.
C) 0.47 J.
D) 1.0 J.

Question

Quizplus · Accepted Answer

The total mechanical energy in a spring-mass system is given by $E = \frac{1}{2} k A^2$. Here, $k = 15 \, \text{N/m}$ and $A = 0.25 \, \text{m}$. Thus, $E = \frac{1}{2} \times 15 \times (0.25)^2 = 0.47 \, \text{J}$.

The Position of a 2 x=Acos⁡(ωt−δ)x = A \cos ( \omega t - \delta )x=Acos(ωt−δ)

The Position of a 2 $x = A \cos ( \omega t - \delta )$