Question

Mollie is using springs to hang art projects. She hang a 3.0kg picture on one particular spring and it stretches from 30 cm to 45 cm long. What is the spring constant for that spring? If she hangs a 2 kg picture beneath the first one, how long will the spring get?

Answer 1

Spring constant, k = 784 N/m

Distance = 49 cm

Step-by-step explanation:

Given:

Mass, m₁ = 3 kg

Distance, x₁ = 30 cm

Distance, x₂ = 45 cm

Spring constant, k = ?

(a)

When the mass is being stretched at a height then potential energy gets converted to kinetic energy.

Potential energy = kinetic energy

`mgh = (1)/(2) k(h-x)^2`

On substituting the value, we get:

`3 X 9.8 X 0.3 = (1)/(2) X k X (0.45 - 0.3)^2\\ \\8.82 = (1)/(2) X k X (0.15)^2\\ \\17.64 = k X 0.0225\\\\k = (17.64)/(0.0225)\\ \\k = 784 N/m`

(b)

Total mass, m = 3 kg + 2kg

= 5 kg

Distance, x = ?

We know:

Force = mg

Force on a spring = kx

`mgh = (1)/(2)k (x - h)^2`

`5 X 9.8 X 0.3 = (1)/(2) X 784 X (x - 0.3)^2\\\\14.7 = 392 (x - 0.3)^2\\\\0.0375 = ( x - 0.3)^2\\\\0.194 = x - 0.3\\\\x = 0.494 m`

The spring stretches to 0.494m or 49 cm when a weight of 2kg is added