Question

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

Answer 1

We can use the concept of Hooke's law to solve this problem. Hooke's law states that the extension of a spring or elastic cord is directly proportional to the stretching force, as long as the elastic limit is not exceeded. Mathematically, we can write:

F = kx

where F is the stretching force, x is the extension, and k is the spring constant (a measure of the stiffness of the cord). To find the length of the cord for a given stretching force, we need to first determine the spring constant.

For the given cord, we know that it can be stretched to its elastic limit by a load of 2N. This means that the maximum stretching force F_max = 2N. At this force, the extension x_max = 0.6cm. We can use these values to find the spring constant:

k = F_max / x_max
k = 2N / 0.6cm
k = 3.33 N/cm

Now we can use this value of k to find the length of the cord for a different stretching force. Let's say we want to find the length of the cord when the stretching force is 2.5N. We can again use Hooke's law:

F = kx

Solving for x, we get:

x = F / k

Plugging in the values, we get:

x = 2.5N / 3.33 N/cm
x = 0.75 cm

Therefore, the length of the cord when the stretching force is 2.5N will be 35 cm + 0.75 cm = 35.75 cm.