Answer :
The spring stretches 0.120 meters from its unstrained length when a 1.5 kg object is suspended from it.
To find the amount by which the spring is stretched from its unstrained length, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, it's expressed as:
[tex]\[ F = -kx \][/tex]
Where:
[tex]- \( F \) is the force exerted by the spring,[/tex]
[tex]- \( k \) is the spring constant,[/tex]
[tex]- \( x \) is the displacement from the equilibrium position.[/tex]
Given that the object's mass is [tex]\( m = 1.5 \) kg and the spring constant is \( k = 122 \) N/m, we can find the displacement (\( x \)).[/tex]
First, let's calculate the force exerted by the spring using the formula:
[tex]\[ F = mg \][/tex]
Where [tex]\( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).[/tex]
[tex]\[ F = (1.5 \, \text{kg}) \times (9.8 \, \text{m/s}^2) \][/tex]
[tex]\[ F = 14.7 \, \text{N} \][/tex]
Now, we can use Hooke's Law to find [tex]\( x \):[/tex]
[tex]\[ F = -kx \][/tex]
[tex]\[ 14.7 \, \text{N} = -(122 \, \text{N/m}) \times x \][/tex]
Now, solve for [tex]\( x \):[/tex]
[tex]\[ x = \frac{14.7 \, \text{N}}{122 \, \text{N/m}} \][/tex]
[tex]\[ x \approx 0.120 \, \text{m} \][/tex]
So, the spring is stretched by approximately [tex]\( 0.120 \, \text{m} \)[/tex] from its unstrained length.
