Answer:
The spring constant of the spring is 180 N/m.
Explanation:
It is given that,
Mass of the block, m = 0.42 kg
The block is pulled so that the spring stretches for 2.1 cm relative to its unstained length.
When the block is released, it moves with an acceleration of [tex]9\ m/s^2[/tex]. We need to find the spring constant of the spring. For a block to be in equilibrium, the net force is given by :
[tex]kx=ma[/tex]
k is spring constant of the spring
[tex]k=\dfrac{ma}{x}\\\\k=\dfrac{0.42\times 9}{2.1\times 10^{-2}}\\\\k=180\ N/m[/tex]
So, the spring constant of the spring is 180 N/m.
