Answer:
The block's speed as it slides horizontally is 3.11 m/s, approximately. As going up on the ramp, it reaches a 0.49 m height before turning back.
Explanation:
As the spring is compressed, it stores elastic potential energy
[tex] U_s = \frac{1}{2} kx^2 [/tex],
as soon as it's released, assuming no loss in the transfer, all the energy is transfered to the block, which adquires kinetic energy
[tex]U_s = K_b \\ \frac{1}{2} kx^2 = \frac{1}{2} m v_b^2 [/tex].
From this last one, we can write for the block's velocity
[tex] v_b = \sqrt{\frac{kx^2}{m}} = \sqrt{\frac{400\frac{N}{m}\times \left(0.220 m\right)^2}{2kg}} \approx \mathbf{3.11 m\s}. [/tex]
Last, when the block reaches the ramp, as it goes up, all its kinetic energy becomes gravitational potential energy, i.e.,
[tex] U_s = K_b = U_b \\ \frac{1}{2}kx^2 = mgh \\ h = \frac{kx^2}{2mg} = \frac{400\frac{N}{m}\times\left(0.220 m\right)^2}{2\times 2 kg\times 9.81 \frac{m}{s^2}} \approx \mathbf{0.49 m}. [/tex]
