How do you find the new length of a spring

A 4.20 kg mass suspended from a spring with spring constant, k =775 N/m, extends it to a total length of 0.270 m. Find the new total length of the spring when a 13.20 kg mass is suspended.

How do you solve this?

Hooke's law in mathematical terms => F = -kΔx, where k is the constant and Δx is the change in extension in metres.

Now the change in your masses is 13.2 - 4.2 = 9 kg. This is what you are adding. F = ma or Weight = mass x g = 9 x 10 = 90 N.

So with F = -kΔx we get 90 = -775Δx
Δx = 90/-775 = -0.116 m

Now you notice your answer is negative, this is because Hooke's law calculates the direction down as a negative and up as positive. But your question has taken the extension downwards as a positive value. You may think taking the - out of the equation F = -kΔx to F = kΔx will make it easier, but this is untrue and may lose you marks depending on the level you are at and subject it is (mechanical maths or physics).

So to get the final answer you do this: 0.270 - -0.116 = 0.386 m.

F=k(x1-x0)
where F = force on spring, x1 is final length of spring, x0 is initial length of spring
F=ma=kx1-kx0
solve for x0
x0 = (kx1 - ma) / k
x0 = (775*0.270 - 4.2*9.81) / 775
x0 = (209.25 - 41.202) / 775
x0 = 0.2168m
now solve F=kx with new x1 and new mass
ma = kx1 - kx0
x1 = (kx0 + ma) / k
x1 = (775*0.217 + 13.2*9.81) / 775
x1 = (168.175 + 129.492) / 775
x1 = 0.384m

Set z = to the length of the unstretched spring
let x1 and x2 equal our lenghts of stretch under m1 and m2
F=kx is your basic equation.
F1 = k ( x1)
F2 = k ( x2)
(z + x1) = 0.270

F1 = k ( x1)
4.2(9.81) = 775 x1
x1 = .053 m
z = 0.270 - 0.053 = 0.217 m

13.20(9.81) = 775x2
x2 = .167 m

total stretched length under m2 is 0.167 + 0.217 = 0.384 m