Question

Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end so that the resultant wire has length 10L. What is the "spring constant" of the resulting wire?

Answer 1

`K_(eq)= (K)/(10)`

Step-by-step explanation:

When springs are connected end to end their effective spring constant

`(1)/(K_(eq))=(1)/(K_1) +(1)/(K_2).................................(1)/(K_(10))`

all ten springs same spring constant K

⇒K=K1=K2........K10

therefore,

`(1)/(K_(eq))=(1)/(K) +(1)/(K).................................(1)/(K)`

`(1)/(K_(eq)) =(10)/(K)`

⇒
`K_(eq)= (K)/(10)`

Answer 2

`K_(system) = (k)/(10)`

Step-by-step explanation:

When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,

`(1)/(K_(system)) = ((1)/(K_(1)))+((1)/(K_(2)))+((1)/(K_(3)))+ ((1)/(K_(4)))+.....+((1)/(K_(n)))`

Here, n = 10

Spring constant of each spring = k

Thus,

`(1)/(K_(system)) = ((1)/(K_(1)))+((1)/(K_(2)))+((1)/(K_(3)))+ ((1)/(K_(4)))+.....+((1)/(K_(10)))`

`(1)/(K_(system)) = ((1)/(k))+((1)/(k))+((1)/(k))+((1)/(k))+((1)/(k))+((1)/(k))+((1)/(k))+((1)/(k))+((1)/(k))+((1)/(k))`

`(1)/(K_(system)) = (10)/(k)`

`K_(system) = (k)/(10)`