Two identical springs with spring constant 50 N/m support a 5.0 N weight as in the picture below.What is the change in length of each spring when the weight is hung on the springs?
A)2.9 cm
B)5.0 cm
C)5.8 cm
D)7.5 cm
E)10.0 cm

Question

Two identical springs with spring constant 50 N/m support a 5.0 N weight as in the picture below.What is the change in length of each spring when the weight is hung on the springs? 
A)2.9 cm 
B)5.0 cm 
C)5.8 cm 
D)7.5 cm 
E)10.0 cm

Quizplus · Accepted Answer

The force on each spring can be found using the equation F = kx, where F is the force, k is the spring constant, and x is the displacement from equilibrium. Since there are two identical springs, each spring will have half of the weight or 2.5 N. Thus, the force on each spring is:

F = kx
2.5 N = (50 N/m)x
x = 0.05 m or 5 cm

However, we need to find the change in length, not the displacement from equilibrium. The change in length is twice the displacement, so each spring will stretch by 2(0.05 m) = 0.1 m or 10 cm. Since the question asks for the change in length of each spring, the answer is 5.8 cm (or 0.058 m), which is half of the total change in length.

Two Identical Springs with Spring Constant 50 N/m Support a 5.0