Question

A block of mass $2kg$ 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 $6m/s$ and $3m/s$, respectively. At this moment the potential energy stored in the spring in increasing at a rate $15J/s$. Find the acceleration of the block at this instant.

• A. $1.5m/s^2$
• B. $3.0m/s^2$
• C. $4.5m/s^2$
• D. $2.5m/s^2$

Answer 1

Answer: (A)

Given, mass of block, $m=2kg$
Since, block is pulled by spring at an instant,
velocity of spring's end, $v_1=6m/s$
and velocity of block end, $v_2=3m/s$
The rate of potential energy,
$\frac{\Delta U}{\Delta t}=\frac{\frac{1}{2}m\left(v_2^2-v_1^2\right)}{t}$
$15=\frac{1}{2}\times 2\frac{\left(v_2+v_1\right)\left(v_2-v_1\right)}{t}$
$\frac{15}{v_2+v_1}=\frac{v_2-v_1}{t}$
Now, acceleration $a=\frac{15}{6+3}$
$a=\frac{15}{9}$
$a=1.6m/s^2$
$\simeq 1.5m/s^2$