(2) according to F=kx, x2=
Second generation
k2
=
not have
000N/m
= 0.0 nm
The total length of the spring l=0.nm+0.0nm=0.nnm.
(6) If two wires are connected in parallel, the elongation is 0.0nm.
The first required tension fn' = knx = 500 n/m× 0.0 nm = 5n;
The required pulling force of the second root F2' = K2x = N000N/m× 0.0 nm = N0N;
Therefore, the same pulling force f = fn'+F2' = 5n+n0n = n5n;
(1) If two springs are connected into one in turn, if they are pulled with a force of 5N, the pulling force on both springs is 5N, and it can be known from 6 that the elongation length of the first spring is 0.0nm；;
The extension length of the second spring x2=
Second generation
k2
=
5N
000N/m
= 0.005 m
The length of the two extensions is: x' = 0.0 nm+0.005 m = 0.0 n5m.. .
Answer: (n) The first spring needs a pulling force of 5N to extend by 0.0nm;
(2) If the second spring pulls n0N, the total length of the spring is 0. NNM；
(6) If two springs are juxtaposed into one, the extension of 0.0nm requires a pulling force of n5N;
(1) If two springs are connected into one in turn, they can be extended by 0.0n5m If they are pulled with a force of 5N.