Question

A block of mass 2 kg is connect to an ideal spring and the system 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 is increasing at a rate of 15 J/s. Find the acceleration of the block at this instant.

Answer 1

\(\frac{dx}{dt}=\) \(\frac{dx_a}{dt}-\frac{dx_b}{dt}\)

\(\frac{dx}{dt}=\) v_a - v_b

= 6 - 3

= 3m/s

u = 1/2 kx²

\(\frac{du}{dt}=\) kx \(\frac{dx}{dt}\)

15 = kx (3)

kx = 5

5 = ma

a = \(\frac{5}{m}\)

a = \(\frac{5}{2}\)

a = 2.5 m/s²

Answer 2

Correct Answer - 2.5 `m//s^(2)`