A spring streatches by 6cm when supporting a load of 15N what load would make the spring extend by 25mm
F = k x
15 = k * 6
k = 15/6
5 = (15/6) x
x = 5 * (6/15 ) = 2 cm
x/25 = 15/6
x = 25/6 * 15 = ____ N
F=ke, thus k=F/e
k=15/0.06
k=250N/m
when e=25mm=0.025mm
then, F=250×0.025
=6.25N
To find the load that would make the spring extend by 25mm, we can use Hooke's Law, which states that the force needed to stretch or compress a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law can be expressed as:
F = k * x
Where:
F is the force applied to the spring,
k is the spring constant, and
x is the displacement of the spring.
In this case, we have the following information:
Initial displacement (x1) = 6cm
Load (F1) = 15N
New displacement (x2) = 25mm = 2.5cm
We need to find the new load (F2).
First, let's calculate the spring constant (k). We can use the initial displacement and the load for this:
k = F1 / x1
k = 15N / 6cm = 2.5 N/cm
Now we can use the spring constant and the new displacement to find the new load:
F2 = k * x2
F2 = 2.5 N/cm * 2.5 cm = 6.25 N
Therefore, a load of 6.25N would make the spring extend by 25mm.