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

Options

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

Answer

A

Explanation

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