A force of 1000 pounds compresses a spring 6 inches from its natural length. How much work (in foot pounds, correct to the nearest whole foot pound) is done in compressing the spring an additional 7 inches?
The text book has the following answers...
a. 88
b. 297
c. 400
d. 566
e. 650
f. 688
g. 924
h. 933
i. 1000
j. none of these
2006-12-03 02:22:38 · 3 answers · asked by Doug 2 in Science & Mathematics ➔ Mathematics
first
k=F/d=1000/6=166.7
k-spring constant
F2=kd2=166.7(6+7)=2167 pounds
The energy stored in the spring = E
correction
E=(kd^2)/2
E=[(166.7)(13)^2]/2= 14,086 foot pounds
So none of the choices are correct.
2006-12-03 02:29:47 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
if the spring constant is k and distance is x
then force = k x
1000 = k 6
k = 1000/6
so for additional 7 inches it is 7*1000/6 = 1166.67 pounds
2006-12-03 10:32:38 · answer #2 · answered by Rajkiran 3 · 0⤊ 0⤋
There is not enough information to solve the problem
2006-12-03 10:35:51 · answer #3 · answered by blue_is_cool2000 1 · 0⤊ 0⤋