What is the distance stretched by a 5-kg barbell?

Question

Hooke's law states that the distance (d) that a spring is stretched by a hanging object varies directly as the weight (w) of the object. If a spring is stretched to 40cm by a 3-kg barbell, what is the distance stretched by a 5-kg barbell?

Answer 1

Hooke's Law:

 

F = -kx

 

Where k is a constant and x is the distance that a spring is stretched from its rest position.  If you hang a barbell from a spring, the force of gravity is pulling on the spring.  The spring will stretch until the forces balance:

 

-kx = mg

 

Where m is the mass of the barbell and g is the acceleration due to gravity (-9.8 m/s²).  For the 3 kg barbell:

 

-k(40) = (3 kg)*(-9.8 m/s²)

 

k = (3)*(9.8)/40 = 0.735

 

For a 5 kg barbell, then:

 

-kx = mg

 

-(0.735)*x = (5 kg)*(-9.8 m/s²)

 

x = (5)*(9.8)/(0.735) = 66.67 cm

 

You can also solve this by noting that since Hooke's Law is linear, the amount of stretching will be proportional to the weight of the barbell; that is a 5 kg weight will stretch the spring 5/3 as far as a 3 kg weight:

 

(3 kg)/(40 cm) = (5 kg)/x cm

 

x = (5/3)*40 = 66.67 cm