What is the increase in elastic potential energy when ghe stretching force in increased by 200 kN?

Options

A. 238.5 J
B. 636.0 J
C. 115.5 J
D. 79.5 J

Answer

b

Explanation

Initial elastic potential energy.
`U_(2) = (1)/(2) F Delta I = (1)/(2) xx (100 xx 1000) x (1.59 xx 10^(-3)) = 79.5 J`
Let `Delta I_(1)` be the elongation in the rod when stretching force is increased by 200 N. Since, `Delta I = (F)/(pi r^(2)) xx (1)/(Y)`, so , `Delta I prop F`
`:. (Delta I_(1))/(Delta I) = (F_(1))/(F) = (100 + 200)/(100) = 3`
or `Delta I_(1) = 3 Delta I = 3 xx 1.59 xx 10^(-3) m = 4.77 xx 10^(-3) m`
final elastic potential energy is
`U_(1) = (1)/(2) F_(1) Delta I_(1) = (1)/(2) xx (300 xx 10^(3)) xx (4.77 xx 10^(-3))`
= 715.5 J
Increases in elastic potential energy
= 715.5 - 79.5 = 636.0 J