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

Answer 1

To find the length of the cord when the stretching force is 2.5N, we can use the concept of Hooke's Law.

Hooke's Law states that the force (F) applied to an elastic material is directly proportional to the extension (x) of the material, as long as the elastic limit is not exceeded.

Mathematically, this can be expressed as: F = k * x, where k is the spring constant.

In this case, when a load of 2N is applied, the cord is stretched to its elastic limit. Let's assume that the original length of the cord is L cm.

Given that a 35cm length of the cord is extended by 0.6cm with a force of 0.5N:
0.5N = k * 0.6cm
k = 0.5N / 0.6cm
k = 0.833 N/cm

Now, let's find the length of the cord when the force is 2.5N.
Using Hooke's Law, we can set up the following equation: 2.5N = 0.833 N/cm * x

Solving for x, we get:
x = 2.5N / 0.833 N/cm
x = 3 cm

Therefore, the length of the cord when the stretching force is 2.5N would be 3 cm.