If 50 ft lbs of work is done on the same spring (in part a) in order to compress it, by how much is the spring shortened?

*Information on part (a) is 400 #/in is the k constant and 1.75" is the compression.

50 ft lbs = 600 in lbs

equation is: 1/2 * 600" lb * 1.75"^2 = 918.75 lbs???

I'm looking for the length in this question, how do I get that?

Ask a New Question or answer this question.

Similar Questions

a spring has a natural length of 20meters.a) find the work required to stretch the spring from 20m to 30m b) from 30m to 40m c)
1. 0 answers
A spring has a spring constant of 25N/m .Part A How much work is required to stretch the spring 2.0cm from its equilibrium
1. 2 answers
A spring has a spring constant of 25N/m .Part A How much work is required to stretch the spring 2.0cm from its equilibrium
1. 1 answer
It takes 6.7 J of work to stretch a spring 4.4 cm from its unstressed (relaxed) length.How much work (in J) is needed to
1. 2 answers