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

To solve the problem, we can use the formula for the spring potential energy: 
PE = (1/2)kx^2 
where PE is the potential energy, k is the spring constant, and x is the displacement (or change in length) of the spring from its equilibrium position.
Since we have two identical springs in parallel, the weight is shared equally between them, so each spring only needs to support half of the weight or 2.5 N. Plugging in the values given, we get: 
PE = (1/2)(50 N/m)(x)^2 = 2.5 N * x 
Solving for x, we get: 
x = sqrt((2.5 N * 2)/(50 N/m)) = 0.158 m = 15.8 cm 
However, this is the total change in length for both springs combined. To find the change in length of each spring, we divide by two: 
Change in length = 15.8 cm / 2 = 7.9 cm 
Therefore, the correct answer is C) 5.8 cm (which is the closest option to 7.9 cm).

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