A spring with a force constant of 5.2 N/m has a relaxed length of 2.45 m. When a mass is attached to the end of the spring and allowed to come rest, the vertical length of the spring is 3.57 m. What is the elastic potential energy stored in the spring?

Question

Sebastian Acosta · Accepted Answer

#2.2"J"#

Hooke's Law states that force #prop# extension:

#F=-kx#

The work done to stretch the spring will be given by force x extension. Because the force varies with extension we need to use some calculus:

#W=kint_0^(x)x.dx=(kx^2)/2#

This will be equal to the elastic potential energy stored in the spring so:

The extension #x=3.57-2.25=0.92"m"#

#:.W=(5.2xx0.92^2)/2=2.2"J"#

Adrian Ayala · Answer

undefined The elastic potential energy stored in the spring can be calculated using the formula: $ PE = \frac{1}{2} k (\Delta L)^2 $, where $ k $ is the force constant of the spring and $ \Delta L $ is the change in length of the spring. In this case, $ k = 5.2 \, \text{N/m} $ and $ \Delta L = 3.57 \, \text{m} - 2.45 \, \text{m} = 1.12 \, \text{m} $. Substituting these values into the formula, we get: $ PE = \frac{1}{2} \times 5.2 \, \text{N/m} \times (1.12 \, \text{m})^2 $. Calculating this gives the elastic potential energy stored in the spring.