Answer to Question #134829 in Physics for Andrew McGuire

Question

Consider a block that is connected to the surface of the Earth with a vertical spring.

You know that the mass of the block is 1.4 kg, the stiffness of the spring is 520.5 N/m, the relaxed length of the spring is 0.7 m, the initial height of the block is 0.4 m, the final height of the block is 1.2 m, and that the block is initially moving vertically downward with a velocity of magnitude 9.9 m/s. What must be the magnitude of the block’s velocity in the final state?

Answer 1

From the conservation of energy:

$$0.5mv^2_i+0.5k(h_i-l)^2+mgh_i\\=0.5mv^2_f+0.5k(h_f-l)^2+mgh_f$$

$$0.5(1.4)(9.9)^2+0.5(520.5)(0.4-0.7)^2+(1.4)(9.8)(0.4)\\=0.5(1.4)v^2_f+0.5(520.5)(1.2-0.7)^2+(1.4)(9.8)(1.2)$$

$$v_f=4.8\frac{m}{s}$$