Question

A block of mass 0.414 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result, the spring is stretched by 0.668 m. Find the spring constant .

____N/m

Answer 1

The spring constant for a spring stretched 0.668 m by a block of mass 0.414 kg at equilibrium is approximately 6.08 N/m, calculated using Hooke's Law with the force due to gravity balancing the spring's force.

Step-by-step explanation:

To find the spring constant (k), we use Hooke's Law, which states that the force (F) exerted by a spring is equal to the spring constant (k) multiplied by the displacement (x) from its equilibrium position, expressed as F = kx. In this scenario, the only force acting on the block when it is in equilibrium is gravity (F = mg), which is counterbalanced by the spring's force.

We have:

• Mass (m) = 0.414 kg
• Gravitational acceleration (g) = 9.81 m/s2
• Displacement (x) = 0.668 m

First, calculate the force due to gravity:

F = mg = 0.414 kg * 9.81 m/s2 = 4.06134 N

Since the spring is in equilibrium, this gravitational force is equal to the elastic force exerted by the spring, which is F = kx. We can rearrange this to solve for the spring constant (k):

k = F / x = 4.06134 N / 0.668 m = 6.08 N/m

So, the spring constant of the spring is approximately 6.08 N/m.