A spring with spring constant k = 800 N/m is compressed 12 cm from its equilibrium position. A spring with spring constant k = 400 N/m has the same elastic potential energy as the first spring when its extension is
A) 0.060 m.
B) 0.085 m.
C) 0.12 m.
D) 0.17 m.
E) 0.24 m.

Question

Quizplus · Accepted Answer

The elastic potential energy of a spring is given by U = (1/2)kx^2, where k is the spring constant and x is the displacement from equilibrium position.

For the first spring with k = 800 N/m and x = 0.12 m, the elastic potential energy is: 
U1 = (1/2)(800 N/m)(0.12 m)^2 = 5.76 J

To find the extension of the second spring with k = 400 N/m that has the same elastic potential energy, we can use the fact that the elastic potential energy is the same when:

U2 = U1 
(1/2)k2x2^2 = (1/2)k1x1^2

where the subscripts 1 and 2 refer to the first and second springs, respectively.

Rearranging for x2, we get: 
x2 = sqrt(k1/k2) * x1 
x2 = sqrt(800 N/m / 400 N/m) * 0.12 m 
x2 = 0.17 m

Therefore, the answer is D) 0.17 m.

A Spring with Spring Constant K = 800 N/m Is