Question

A spring hangs from the ceiling with an unstretched length of x 0 = 0.39 m x0=0.39 m . A m 1 = 6.7 kg m1=6.7 kg block is hung from the spring, causing the spring to stretch to a length x 1 = 0.64 m x1=0.64 m . Find the length x 2 x2 of the spring when a m 2 = 3.7 kg m2=3.7 kg block is hung from the spring. For both cases, all vibrations of the spring are allowed to settle down before any measurements are made.

Answer 1

Answer:

The length is 0.55 meters

Explanation:

When mass m1 is hung from the spring, the spring stretches and we wait until the vibrations are settle down, so in this point the mass is at rest and we can use Newton's first law, this is, the net force on the mass is zero so elastic force (Fk) and weight are equal in magnitude:

$W=F_{k}$

weight is mass times gravitational acceleration (g) and elastic force is the length the spring stretches times spring constant (k):

$mg=k(x-x_0)$ (1)

solving for k and using data provided on 1:

$k=\frac{m_1g}{x_1-x_0}=\frac{6.7*9.81}{0.64-0.39}$

$k=262.9\frac{N}{m}$

Now knowing k, we can use equation (1) to find x1:

$x_2=\frac{m_2g}{k}-x_0 =\frac{3.7*9.81}{262.9}+0.39$

$x_2=0.55 m$

Answer 2

The new length of the spring at the given extension is 0.528 m.

The given parameters:

• unstretched length, x₀ = 0.39 m
• stretched length, x₁ = 0.64 m
• mass on the spring, m = 6.7 kg

The spring constant of the spring is calculated as follows;

$F = kx = mg\\\\k = \frac{mg}{x} \\\\k = \frac{6.7 \times 9.8}{0.64 - 0.39} \\\\k = 262.64 \ N/m$

The extension of the spring when 3.7 kg mass is hung on it;

$x = \frac{mg}{k} \\\\x = \frac{3.7 \times 9.8}{262.64} \\\\x = 0.138 \ m$

The new length of the spring at the given extension is calculated as follows;

$x_2 = x+ x_0\\\\x_2 = 0.138 + 0.39\\\\x_2 = 0.528 \ m$

Thus, the new length of the spring at the given extension is 0.528 m.

Learn more here:https://brainly.com/question/4404276