An elastic cord can be stretched to its elastic limit by load of 2N. If a 35cm length of the cord is extended 0.6cm by a force of 0.5N, what will be the length of the cord when the stretching force is 2.5N???

Answer 1

We can use Hooke's Law to solve this problem, which states:

F = kx

where F is the force, k is the spring constant, and x is the extension.

We're given that a 35 cm length of the cord is extended by 0.6 cm by a force of 0.5 N, so we can plug these values into Hooke's Law:

0.5 N = k * 0.6 cm

Now we can find the spring constant (k):

k = 0.5 N / 0.6 cm = 5/6 N/cm ≈ 0.833 N/cm

Now that we have the spring constant value, we can use it to find the extension when the force is 2.5 N:

2.5 N = 0.833 N/cm * x
x = 3 cm

So, the cord extends 3 cm when the force is 2.5 N. To find the total length of the cord including this extension, we just add it to the original length of 35 cm:

Total length = Original length + Extension
Total length = 35 cm + 3 cm = 38 cm

So, the length of the cord when the stretching force is 2.5 N will be 38 cm.