If you need 6 joules of work to expand a spring from 10 to 12 cm and another 10J to expand it from 12 to 14 cm. Wich is the natural lenght of the spring?

1 Answer
Ananda Dasgupta
May 1, 2018

#8\ "cm"#

Explanation:

Let the natural length of the spring and its spring constant be #l_0# and #k# respectively. Then, the potential energy of the spring when its length is #l# is given by

#V = 1/2k(l-l_0)^2#

The work done to change the length of the spring is equal to the increase in its potential energy.

So, we have

#1/2k[(12\ "cm"-l_0)^2-(10\ "cm"-l_0)^2] = 6\ "J"#

and

#1/2k[(14\ "cm"-l_0)^2-(12\ "cm"-l_0)^2] = 10\ "J"#

Dividing these eliminates #k# to give

#[(14\ "cm"-l_0)^2-(12\ "cm"-l_0)^2]/[(12\ "cm"-l_0)^2-(10\ "cm"-l_0)^2] = 10/6 = 5/3 implies#

# [(14^2-12^2)"cm"^2-2times(14\ "cm"-12\ "cm")times l_0]/ [(12^2-10^2)"cm"^2-2times(12\ "cm"-10\ "cm")times l_0] = 5/3implies#

# [2times 26\ "cm"-4l_0]/[2times 22\ "cm"-4l_0]=5/3 implies [13\ "cm"-l_0]/[11\ "cm"-l_0] =5/3implies#

# 39\ "cm"-3l_0=55\ "cm"-5l_0quad implies quad 2l_0=16\ "cm"#

Hence

#l_0=8\ "cm"#

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