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(D) Let $F$ be the tension in the spring. The potential energy $U$ of the spring is given by $U = \frac{1}{2} k x^2$, where $x$ is the extension of th

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$2.5$

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(D) Let $F$ be the tension in the spring. The potential energy $U$ of the spring is given by $U = \frac{1}{2} k x^2$, where $x$ is the extension of the spring.The rate of change of potential energy is $\frac{dU}{dt} = kx \frac{dx}{dt} = F \cdot v_{rel}$, where $v_{rel}$ is the rate of change of extension, which is the relative velocity of the ends of the spring.Here, $v_{rel} = v_A - v_{block} = 6 \,m/s - 3 \,m/s = 3 \,m/s$.Given $\frac{dU}{dt} = 15 \,J/s$, we have $15 = F \cdot 3$, which gives $F = 5 \,N$.The force $F$ is the net force acting on the block of mass $m = 2 \,kg$.Using Newton's second law, $F = ma$, we get $5 = 2 \cdot a$.Therefore, $a = \frac{5}{2} = 2.5 \,m/s^2$.

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