Question

A force of 60 N is used to stretch two springs that are initially the same length. Spring A has a spring constant of 4 N/m, and spring B has a spring constant of 5 N/m.

Answer 1

I guess the problem is asking for the elongation of the two springs.

The relationship between force (F), spring constant (k) and eleongation of the spring with respect to the rest position (x) is given by

`F=kx`

Rearranging, we can find the elongation as:

`x= (F)/(k)`

For spring A, F=60 N and k=4 N/m, therefore the elongation is

`x_A = (60 N)/(4 N/m)=15 m`

For spring B, F=60 N and k=5 N/m, therefore the elongation is

`x_B= (60 N)/(5 N/m)=12 m`