Question

a block of mass 0.25Kg is placed on top of light vertical spring of force constant 500N/m pushed down so that the spring is compressed by 0.10m after the block is released from rest it travels upward and leaves the spring what is the maximum height

Answer 1

Let's first find the velocity of the block after it released from the law of conservation of energy:

$$PE=KE,$$ $$\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2,$$ $$v=\dfrac{kx^2}{m}=\dfrac{500\ \dfrac{N}{m}\cdot(0.1\ m)^2}{0.25\ kg}=20\ \dfrac{m}{s^2}.$$

Let's find the time that the block takes to reach the maximum height:

$$v_y=v_{0}-gt,$$ $$0=v_{0}-gt$$ $$t=\dfrac{v_0}{g}=\dfrac{20\ \dfrac{m}{s}}{9.8\ \dfrac{m}{s^2}}=2.04\ s.$$

Finally, we can find the maximum height reached by the block:

$$y_{max}=v_0t-\dfrac{1}{2}gt^2,$$ $$y_{max}=20\ \dfrac{m}{s}\cdot2.04\ s-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(2.04\ s)^2=20.4\ m.$$