Question

A vertical spring has a spring constant of 100. Newtons per meter. When an object is attached to the bottom of the spring, the spring changes from its unstretched length of 0. 50 meter to a length of 0. 65 meter. The magnitude of the weight of the attached object is.

Answer 1

Hi there!

We can use Newton's Second Law to solve.

Recall the force from a spring:

`F = -k\Delta x`

F = Force (N)
k = Spring constant (N/m)

Δx = displacement from equilibrium (m)

Sum the forces acting on the mass. Weight (Mg) acts downward, while the spring's force is upward.

`\Sigma F_(y) = W - k\Delta x`

Since the mass is in equilibrium at this point, ∑F = 0 N.

`0 = W- k\Delta x\\ \\ k\Delta x = W`

Plug in the givens to solve for weight:

`100(0.65 - 0.50) = \boxed{15 N}`