Question

(c8p7_5e) A spring with a spring constant of 3250 N/m is initially stretched until the elastic potential energy is 1.50 J. (U0. for no stretch.) What is the change in the elastic potential energy if the initial stretch is changed to a stretch of 2.0 cm? Submit Answer Tries 0/1:2 What is the change in the elastic potential energy if the initial stretch is changed to a compression of 2.0 cm? Submit Answer Tries 0/12 What is the change in the elastic potential energy if the initial stretch is changed to a compression of 4.0 cm? Submit Answer Tries 0/12

Answer 1

Initial potential energy of spring is

U_i = 1.50 J

Elastic potential energy of a spring is given by

$U = \frac{1}{2}kx^{2}$

For a stretch of 2 cm

x = 2 cm = 0.02 m

$U = \frac{1}{2}(3250)(0.02)^{2}$

U = 0.65 J

Change in potential energy = U - U_i

Change in potential energy = 0.65 - 1.50

Change in potential energy = -0.85 J

(b) If we compress it by 2 cm , then the energy will be same as for extension of 2cm

x = 2 cm = 0.02 m

$U = \frac{1}{2}(3250)(0.02)^{2}$

U = 0.65 J

Change in potential energy = U - U_i

Change in potential energy = 0.65 - 1.50

Change in potential energy = -0.85 J

(c)

If we compress it by 4 cm

x = 4 cm = 0.04 m

$U = \frac{1}{2}(3250)(0.04)^{2}$

U = 2.6 J

Change in potential energy = U - U_i

Change in potential energy = 2.6 - 1.50

Change in potential energy = 1.10 J