Two identical springs with spring constant 50 N/m support a 5.0 N weight as in the picture below.What is the tension in spring A?
A)1.45 N
B)2.50 N
C)2.89 N
D)3.75 N
E)5.00 N

Question

Two identical springs with spring constant 50 N/m support a 5.0 N weight as in the picture below.What is the tension in spring A? 
A)1.45 N 
B)2.50 N 
C)2.89 N 
D)3.75 N 
E)5.00 N

Quizplus · Accepted Answer

The force exerted by the weight is given by F = mg, where g is the acceleration due to gravity and m is the mass of the weight. In this case, F = 5.0 N.

The force exerted by each spring can be calculated using Hooke's Law:

F = kx

where k is the spring constant and x is the displacement from the equilibrium position. Since the springs are identical, they will undergo equal displacements, so we can divide the weight force evenly between them:

F_spring = F/2 = 2.5 N

Now, we can solve for the displacement:

2.5 N = kx
x = 0.05 m

The tension in spring A is equal to the force exerted by spring A, which is given by:

F_A = kx

We can solve for k using either spring, since they are identical:

2.5 N = k(0.05 m)
k = 50 N/m

Therefore, the tension in spring A is:

F_A = kx = (50 N/m) (0.025 m) = 2.5 N * 0.5 = 1.25 N

However, we have to remember that the weight contributes an additional force, so the total tension in spring A is:

T_A = F_A + F/2 = 1.25 N + 2.5 N = 3.75 N

Therefore, the correct answer is C) 2.89 N.

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