The force required to stretch a Hooke's-law
spring varies from 0 N to 23.1 N as we stretch
the spring by moving one end 10.6 cm from
its unstressed position.
Find the force constant of the spring. An-
swer in units of N/m.
part 2
Find the work done in stretching the spring.
Answer in units of J.
2007-01-11 09:41:31 · 2 answers · asked by kavita 1 in Science & Mathematics ➔ Physics
1)
k = 23.1/0.106 N/m
2)
. . . . . . . . . . . . 10.6
W = (23.1/0.106)∫xdx
. . . . . . . . . . . . 0
W = (23.1/0.106)((0.106)(0.106)/2 - 0) J
2007-01-11 11:06:50 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
F=kx
x=0 ; F=k(0)=0
x=10.6 cm (1m / 100cm)=.106m; F=23.1=k(.106) --> k=(23.1/.106) N/m
W=.5kx^2
x=.106; k=(23.1/.106); W=.5(23.1/.106)(.106)^2 J
No calculator on me...
2007-01-11 18:31:30 · answer #2 · answered by krnxblizzard 2 · 0⤊ 0⤋