Answer to Question #252978 in Mechanics | Relativity for Nana

Question

A vertical spring (ignore its mass), whose spring stiffness constant is 950 N m^-1

is attached to a table. The spring is compressed down 15 cm and a 300 g ball

placed on it. The spring is then released and the ball flied off vertically. Determine the maximum height reached by the ball.

Answer 1

Let's first find the initial velocity of the ball when the spring is released from the law of conservation of energy:

$$PE_{spring}=KE_{ball},$$ $$\dfrac{1}{2}kx^2=\dfrac{1}{2}mv_0^2$$ $$v_0=\sqrt{\dfrac{kx^2}{m}}=\sqrt{\dfrac{950\ \dfrac{N}{m}\cdot(0.15\ m)^2}{0.3\ kg}}=8.44\ \dfrac{m}{s}.$$

Finally, we can fina the maximum height reached by the ball from the kinematic equation:

$$v^2=v_0^2-2gh_{max},$$ $$0=v_0^2-2gh_{max},$$ $$h_{max}=\dfrac{v_0^2}{2g}=\dfrac{(8.44\ \dfrac{m}{s})^2}{2\cdot9.8\ \dfrac{m}{s^2}}=3.63\ m.$$