A block A of mass `m=5kg` is attached with a spring having force constant `K=2000Nm^-1`. The other end of the spring is fixed to a rough plane, inclined at `37^@` with horizontal and having coefficient of friction `m=0.25`. Block A is gently placed on the plane such that the spring has no tension. Then block A is released slowly. Calculate elongation of the spring when equilibrium is achieved. Now an inextensible thread is connected with block A and passes below pulley C and over pulley D, as shown in figure. Other end of the thread is connected with another block B of mass `3kg`. Block B is resting over a table and thread is loose. If the table collapse suddenly and B falls freely through `80//9cm`, the thread becomes taut, calculate combined speed of blocks at that instant, and maximum elongation of spring in the process of motion. `(g=10ms^-2)`.

Explanation

Since the block is released slowly, therefore, it starts to slide down the plane till equilibrium of forces is achieved.

Let at that instant elongation of spring be `x_0`, then tension in it is
`T_0=Kx_0=2000x_0`
Considering free body diagram of block,
`N=mg cos 37^@-40N` (i)
`Kx_0+muN=mgsin37^@` (ii)
From Eq. (i) and (ii),
`x_0=0.01 m=1cm`
When table collapses, first block B falls freely under gravity through height `80//9`. Therefore, its speed just before the string becomes taut is
`v_0=sqrt(2gxx(0.80/9))=4/3ms^-1`
Now block A is jerked into motion and a large tension (for a very small time interval) is developed in string due to which both the blocks A and B experience numerically equal impulses.
Let its magnitude be J and let the combined speed of blocks be v.
Then for block A, `J=5v` (iii)
for block B, `3v_0-J=3v` (iv)
From equation (iii) and (iv) `v=0.5ms^-1`
At the instant of maximum elongation of spring, blocks are momentarily at rest.
Let distance moved by the blocks be x from the instant when blocks A was jerked into motion to the instant of maximum elongation of the spring.
According to law of consevation of energy,
Loss of potential energy of A+Loss of potential energy of B+
Loss of kinetic energy of blocks=Increase in energy stored in spring+Work done by the block A against friction.
`=5.g(x sin 37^@)+3gx+{1/2xx(5v^2)+1/2xx(3v^2)}`
`={1/2K(x_0+x)^2-1/2Kx_0^2}+muNx`
`x=0.5mm` or `5cm`
Maximum elongation of spring= Its initial elongation `(x_0)+`
Further elongation`(x)=6cm`