A load of 4N extends a spring by 10mm. What load would extend it by 15mm?

Answer 1

F= k x
4 Newtons = k* 10 mm
k = 0.40 N/mm
F = 0.40 N/mm * 15mm
F = 6 N

Answer 2

Load 1= 4N
Extension 1= 10mm
Load 2 = ?
Extension 2 = 15mm
Then,
Load=(4N×15mm)/10mm
=6N

Answer 3

Yes, that's another way to solve the problem using Hooke's law, which states that the force (load) applied to a spring is proportional to the extension of the spring. So if you know the constant of proportionality (spring constant), you can use it to find the force required to produce a certain extension. In this case, we have:

F1 = kx1 (where F1 = 4 N and x1 = 10 mm)
F2 = kx2 (where we need to find F2 for x2 = 15 mm)

Dividing the second equation by the first one, we get:

F2/F1 = x2/x1

Solving for F2, we get:

F2 = F1 * x2/x1 = 4 N * 15/10 = 6 N

So both methods lead to the same result, which is the force required to extend the spring by 15 mm.