Question

An object with mass 500g less hang from the spring with force constant 20N/m.How far in cm would the spring stretch?

Answer 1

To calculate the distance the spring stretches, we will use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. Mathematically, this can be written as:

F = -kx

Where F is the force exerted by the spring, k is the force constant, and x is the displacement from the equilibrium position.

In this case, the force exerted by the spring is the weight of the object hanging from it, given by:

F = mg

Where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Given that the mass of the object is 500g (or 0.5kg) less than the weight, we can write the equation as:

F = (m - 0.5g)g

Now, setting this equal to the force exerted by the spring and solving for x:

-kx = (m - 0.5g)g

x = -(m - 0.5g)g / k

Substituting the given values:

m = 0.5 kg
g = 9.8 m/s^2
k = 20 N/m

x = -(0.5 - 0.5(9.8)) * 9.8 / 20

Simplifying the expression:

x = -(0.5 - 4.9) * 9.8 / 20

x = -(4.4) * 9.8 / 20

x = -43.12 / 20

x = -2.156 m

Converting to centimeters:

x = -2.156 m * 100 cm/m

x ≈ -215.6 cm

Therefore, the spring would stretch approximately 215.6 cm.