Question

a light spring with spring constant $k_0$ hangs vertically, as shown on the left side of the figure. a block of mass $m_0$ is then attached to the spring and pulled downward until the spring is displaced a distance $h_0$, as shown on the right side of the figure. the block is released from rest and accelerates upward. what is the acceleration of the block immediately after it is released?
a $g$
b $\frac{k_0h_0}{m_0}$
c $\frac{k_0h_0 + m_0g}{m_0}$
d $\frac{k_0h_0 - m_0g}{m_0}$

Explanation:

Step1: Identify forces on the block

The upward - acting force on the block is the spring force $F_s=k_0h_0$ (by Hooke's law $F = kx$, where $x = h_0$) and the downward - acting force is the gravitational force $F_g=m_0g$.

Step2: Apply Newton's second law

Newton's second law is $F_{net}=ma$. The net force on the block $F_{net}=k_0h_0 - m_0g$. So, $k_0h_0 - m_0g=m_0a$.

Step3: Solve for acceleration

We can solve the equation $k_0h_0 - m_0g=m_0a$ for $a$. First, divide both sides of the equation by $m_0$: $a=\frac{k_0h_0 - m_0g}{m_0}$.

Answer:

D. $\frac{k_0h_0 - m_0g}{m_0}$