Question

[100 points] A force of 6 lb is required to hold a spring stretched 2 in. beyond its natural length. How much work W is done in stretching it from its natural length to 4 in. beyond its natural length?

W = ___________ ft-lb

Answer 1

I'll give you an example and plug in the numbers:

Given:

6 lb is needed to stretch 4 inches beyond natural length.

Need work done to stretch same string from natural length to 8 inches.

Solution:

string stiffness, K

= Force / stretched distance

= 6 lb / 4 inches

= 1.5 lb/inch

Work done on a string of stiffness K

= (Kx^2)/2 lb-in

= 1.5 lb/in *(8 in)^2)/2

= 48 lb-in.

Answer 2

its 48 lb·in.