Question

In the produce section of a supermarket, five pears are placed on a spring scale. The placement of the pears stretches the spring and causes the dial to move from zero to a reading of 2.0 kg. If the spring constant is 450 N/m, what is the displacement of the spring due to the weight of the pears

Answer 1

Answer:

The displacement of the spring due to weight is 0.043 m

Explanation:

Given :

Mass $m = 2$ Kg

Spring constant $k = 450 \frac{N}{m}$

According to the hooke's law,

  $F = -kx$

Where $F =$ force, $x =$ displacement

Here,

$F = mg$         ( $g = 9.8 \frac{m}{s^{2} }$ )

$F = 2 \times 9.8 = 19.6$ N

Now for finding displacement,

  $x = \frac{F}{k}$

Here minus sign only represent the direction so we take magnitude of it.

  $x = \frac{19.6}{450}$

  $x = 0.043$ m

Therefore, the displacement of the spring due to weight is 0.043 m