An elastic cord can be stretched to it’s elastic limit by a load of 2N if a 35cm length of time cord is extended by 0.6cm by a force of0.5N what will be the length of the cord when the stretching force is 2.5N
To solve this question, we need to use Hooke's Law, which states that the extension of an elastic material is directly proportional to the force applied to it within its elastic limit.
Let's denote the original length of the cord as L0 and the original stretching force as F0. Given that the cord can be stretched to its elastic limit by a load of 2N, we can say that L0 + 0.6 cm will be the length of the cord at the elastic limit.
We can now set up a proportion to find the constant of proportionality (k) for this elastic cord:
(F0 / L0) = (2N / 0.6 cm)
To find the value of F0, we need to rearrange the equation:
F0 = (2N / 0.6 cm) * L0
Now, let's solve for L1, which will be the length of the cord when the stretching force is 2.5N. We can use the same proportionality constant (k):
(2.5N / L1) = (F0 / L0)
Rearranging the equation to find L1:
L1 = (F0 / L0) * (2.5N)
Plugging in the value of F0 from the previous equation:
L1 = [(2N / 0.6 cm) * L0 / L0] * (2.5N)
Simplifying the expression:
L1 = (2N / 0.6 cm) * (2.5N)
L1 = (5/3) * 2.5N
L1 = 12.5/3 cm
Therefore, the length of the cord when the stretching force is 2.5N will be approximately 4.17 cm.