It takes 3.38 J of work to stretch a Hooke’s-law
spring 5.63 cm from its unstressed length.
How much the extra work is required to
stretch it an additional 4.22 cm?
Answer in units of J
science - drwls, Saturday, November 17, 2012 at 8:08am
To stretch 4.22+5.63 = 9.85 cm, that is
(9.85/5.63) = 1.750 times the initial stretch amount. The energy required is the square of that ratio (3.06 times) higher than before.
Energy required went from 3.38 J to 10.35 J
The extra work is the difference of thse numbers.