Question

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 to 3.06 10-2 N. The length and radius of the collagen are, respectively, 2.7 and 0.093 cm, and Young's modulus is 3.10 106 N/m2.
(a) If the stretching obeys Hooke's law, what is the spring constant k for collagen?(b) How much work is done by the variable force that stretches the collagen?

Answer 1

Answer:

(a) $k=311.8N/m$

(b) $W=1.501*10^{-6}J$

Explanation:

For Part (a)

Solving for spring constant k

$k=\frac{YA}{L_{o} } \\k=\frac{Y(\pi r^{2} )}{L_{o} }\\k=\frac{(3.10*10^{6} N/m^{2} )(\pi (0.093*10^{-2}m )^{2} )}{2.7*10^{-2}m }\\k=311.8N/m$

For Part (b)

Solving for work is done

$x=\frac{F}{k}\\ x=\frac{3.06*10^{-2}N }{311.8N/m}\\x=9.814*10^{-5}m$

The Work done is given as:

$W=\frac{1}{2}Fx\\ W=\frac{1}{2}(3.06*10^{-2}N )(9.814*10^{-5}m )\\W=1.501*10^{-6}J$