Question

An angler hangs a 4.20-kg fish from a vertical steel wire 2.55 m long and 5.00×10−3cm2 in cross-sectional area. The upper end of the wire is securely fastened to a support.

A) Calculate the amount the wire is stretched by the hanging fish.

Answer:

Δl =	1.05×10−3 m

B) The angler now applies a varying force F⃗  at the lower end of the wire, pulling it very slowly downward by 0.500mm from its equilibrium position. For this downward motion, calculate the work done by gravity.

Answer:

W =	2.06×10−2 J

C) For this downward motion, calculate the work done by the force F⃗ .

Answer:

W =	4.90×10−3 J

D) Calculate the work done by the force the wire exerts on the fish.

Answer:

W =	−2.55×10−2 J

E) Calculate the the change in the elastic potential energy (the potential energy associated with the tensile stress in the wire).

Answer: ??? I can't figure out this answer.

Answer 1

(A)

Amount the wire is stretched by the hanging fish,

l = F*l / Y*A

Young's modulus for steel = 20*10^10 Pa

l = 4.20*9.81*2.55 / 20*10¹⁰*5*10^-3*10^-4

l = 1.05*10^-3 m

(B)

work done by gravity,

W = mg*x

W = 4.20*9.81*0.50*10^-3

W = 0.0206 J

(C)

work done by the force F,

W = Y*x^2*A / 2l

W = 20*10¹⁰*(0.50*10^-3)²*5*10^-7 / 2*2.55

W = 4.90*10^-3 J

(D)

Work done by the force the wire exerts on the fish,

W = -YAl'*x / l = -YA(l + x / 2)*x / l

W =  20*10¹⁰*5*10^-7(1.05*10^-3 + 0.5*10^-3 / 2)*0.5*10^-3 / 2.55

W = -0.0255 J

(E)

change in the elastic potential energy,

U = (1/2)*(YA / l)*[(l + x)^2 - l^2]

U = 20*10¹⁰*5*10^-7[(1.05*10^-3 + 0.5*10^-3)^2 - ( 0.5*10^-3)^2] / 2*2.55

U = 0.0422 J

U = 4.22*10^-2 J