Question

a body of mass m=2 kg is hung from a spring constant k=25N/m as shown in the figure. in equilibrium, the elongation in the spring will be​

Answer 1

Given:

A body of mass m=2 kg is hung from a spring constant k=25N/m as shown in the figure.

To find:

Elongation in the spring at equilibrium?

Calculation:

When the mass is at equilibrium, the weight of the object will be equal and opposite to the spring force.

$\therefore \sf \: weight = spring \: force$

$\sf \implies \: mg = kx$

$\sf \implies \: 2 \times 10 = 25x$

$\sf \implies \: 25x = 20$

$\sf \implies \: x = \dfrac{20}{25}$

$\sf \implies \: x = 0.8 \: m$

So, the elongation on the spring will be 0.8 metres.