Question

a spring of spring constant k=12.5N/m is hung vertically. a 0.500kg mass is then suspended from the spring. what is the displacement of the end of the spring due to the wieght of the 0.500 kg mass?

Answer 1

Answer:

The displacement of one end of the spring due to weight = 0.4 m.

Explanation:

Given :

Spring constant $(k) = 12.5 \frac{N}{m}$

Mass $(m) = 0.5 kg$

From the hooke's law,

Force is proportional to the negative of the displacement.

  $F$ ∝ $-x$

  $F = -kx$

Where minus sign represent that, the force is always directed toward mean position.

Now one force is gravity $mg$ acts as a downward. so we write,

  $mg = kx$

Here we take only magnitude.

    $x = \frac{mg}{k}$

Where $x =$ displacement of one end of spring, $g = 10 \frac{m}{s^{2} }$

    $x = \frac{0.5\times 10}{12.5}$

    $x = 0.4$ m