Asked by david
a force of 1250 pounds compresses a spring 5 inches from its natural length. find the work done in compressing the spring 8 additional inches.
f=kd
1250=k5
k=250
then what do i do.
the answer choices are:
a.3250
b.21,125
c.18000
d.2000
e. none of these
Work = (1/2)* F * deflection
After 8 additional inches of deflection, the Force will increase to
(13/5)*1250 = 3250 lb
The total work done (from zero deflection) will be
(1/2)*3250 *13 = 21,125 in-lb
The original work done (for the first 5 inches of deflection) was
(1/2)*1250*5 = 3125 in-lb
The additional work done is
21,125 - 3125 = 18,000 in-lb
Another way to do this would be to calculate the spring constant k = 250 lb/in, as you have done. Then subtract
(1/2) k 5^2 from (1/2) k (13^2).
The difference is (1/2)*k * 144 = ?
f=kd
1250=k5
k=250
then what do i do.
the answer choices are:
a.3250
b.21,125
c.18000
d.2000
e. none of these
Work = (1/2)* F * deflection
After 8 additional inches of deflection, the Force will increase to
(13/5)*1250 = 3250 lb
The total work done (from zero deflection) will be
(1/2)*3250 *13 = 21,125 in-lb
The original work done (for the first 5 inches of deflection) was
(1/2)*1250*5 = 3125 in-lb
The additional work done is
21,125 - 3125 = 18,000 in-lb
Another way to do this would be to calculate the spring constant k = 250 lb/in, as you have done. Then subtract
(1/2) k 5^2 from (1/2) k (13^2).
The difference is (1/2)*k * 144 = ?
Answers
Answered by
B
work= force * distance, so you can set up an integral to solve for work
F=k*x, k=250
force =250x
distance =dx
so then you integrate(250x)dx from 0->(5+8)=21,125
idk if its right but its an answer choice
F=k*x, k=250
force =250x
distance =dx
so then you integrate(250x)dx from 0->(5+8)=21,125
idk if its right but its an answer choice
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