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A block of mass $2 \mathrm{~kg}$ is connected to an ideal spring and is placed on a smooth horizontal surface. The spring is pulled to move the block and at an instant, the speed of end $A$ of the spring and speed of the block were measured to be $6 \mathrm{~m} / \mathrm{s}$ and $3 \mathrm{~m} / \mathrm{s}$, respectively. At this moment the potential energy stored in the spring in increasing at a rate $15 \mathrm{~J} / \mathrm{s}$. Find the acceleration of the block at this instant.
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Verified Answer
The correct answer is:
$1.5 \mathrm{~m} / \mathrm{s}^2$
Given, mass of block, $m=2 \mathrm{~kg}$
Since, block is pulled by spring at an instant, velocity of spring's end, $v_1=6 \mathrm{~m} / \mathrm{s}$ and velocity of block end, $v_2=3 \mathrm{~m} / \mathrm{s}$
The rate of potential energy,
$$
\begin{aligned}
\frac{\Delta U}{\Delta t} & =\frac{\frac{1}{2} m\left(v_2^2-v_1^2\right)}{t} \\
15 & =\frac{1}{2} \times 2 \frac{\left(v_2+v_1\right)\left(v_2-v_1\right)}{t} \\
\frac{15}{v_2+v_1} & =\frac{v_2-v_1}{t}
\end{aligned}
$$
Now, acceleration $a=\frac{15}{6+3}$
$$
\begin{aligned}
a & =\frac{15}{9} \\
a & =1.6 \mathrm{~m} / \mathrm{s}^2 \\
& \simeq 1.5 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
Since, block is pulled by spring at an instant, velocity of spring's end, $v_1=6 \mathrm{~m} / \mathrm{s}$ and velocity of block end, $v_2=3 \mathrm{~m} / \mathrm{s}$
The rate of potential energy,
$$
\begin{aligned}
\frac{\Delta U}{\Delta t} & =\frac{\frac{1}{2} m\left(v_2^2-v_1^2\right)}{t} \\
15 & =\frac{1}{2} \times 2 \frac{\left(v_2+v_1\right)\left(v_2-v_1\right)}{t} \\
\frac{15}{v_2+v_1} & =\frac{v_2-v_1}{t}
\end{aligned}
$$
Now, acceleration $a=\frac{15}{6+3}$
$$
\begin{aligned}
a & =\frac{15}{9} \\
a & =1.6 \mathrm{~m} / \mathrm{s}^2 \\
& \simeq 1.5 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
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