Answer:
The spring is compressed 0.11 m.
Explanation:
We will use the conservation of energy between the initial position (A) and the position when the block comes to rest (B).
[tex]K_A + U_A = K_B + U_B\\0 + mgh = 0 + \frac{1}{2}kx^2\\h = 3\times \sin(35^\circ) = 1.72 m\\\\12\times 9.8 \times 1.72 = \frac{1}{2}(3\times 10^4)x^2\\x = \sqrt{\frac{2*12*9.8*1.72}{3\times 10^4}}\\x = 0.11 ~m[/tex]
Since the block is released from rest, and comes to rest finally, first and final kinetic energies are zero.
since there is no friction, the initial gravitational potential energy of the block is equal to the final elastic potential energy stored in the spring.
