An elastic cord can be stretched to its elastic limit by a load of 2 N . if a 35cm length of the cord is extended 0.6 cm by a force of 0.5N what will be the length of the cord when the stretching force is 2.5N
Well of course if it breaks at over 2 N it is hard to say but if it just stops stretching then use 2 N
2 N / .5 N = 4
so since x = F/k
use 4 * .6 = 2.4 cm
To solve this problem, we can use the concept of Hooke's Law, which states that the extension of a spring (or elastic cord) is directly proportional to the applied force.
First, let's find the spring constant (k) of the elastic cord. The spring constant represents the stiffness of the cord and can be calculated using Hooke's Law formula:
k = F / x
where k is the spring constant, F is the force applied, and x is the extension of the cord.
Given that the load of 2 N results in an extension of 35 cm - 0.6 cm = 34.4 cm:
k = 2 N / (34.4 cm) = 0.0581 N/cm
Now, we can use the spring constant to find the extension of the cord when the stretching force is 2.5 N. Let's denote the new extension as x2.
x2 = F2 / k
where F2 is the stretching force and k is the spring constant.
Plugging in the values, we get:
x2 = 2.5 N / (0.0581 N/cm)
Simplifying, we find:
x2 = 43.06 cm
Therefore, the length of the cord when the stretching force is 2.5 N will be 35 cm + 43.06 cm = 78.06 cm (rounded to two decimal places).