F = kx, so
x/10 = .01/2
where x is the amount of extension.
A force of 5n applied to a felastic spring 2m makes it to extendly 0.01 what will be the new length if force is 10n is applied
2 answers
delta x is the amount of extension
5 = k (.01)
k = 5/.01 = 500 N/m
10 = 500 (delta X)
delta x = 1/50 = .02 meter
so new length = 2,02 meters
5 = k (.01)
k = 5/.01 = 500 N/m
10 = 500 (delta X)
delta x = 1/50 = .02 meter
so new length = 2,02 meters