An elastic cord can be streached to its elastic limit by a load of 2N. If 35cm lenght of the cord is extended 0.6cm by force of 0.5N what will be the lenght of the cord?
Let's denote the original length of the cord as L, the extension under the force of 0.5N as x, and the new length of the cord as L'.
According to Hooke's Law, the extension of an elastic material is directly proportional to the force applied to it, as long as the elastic limit is not reached. Therefore, we can write:
x = (F / k)
Where:
x = extension (0.6cm)
F = force applied (0.5N)
k = spring constant
Given that the elastic limit is reached with a load of 2N, we can find the spring constant k:
k = F / x
k = 0.5 / 0.6
k = 0.8333 N/cm
Now, we can find the new length of the cord when the force is 0.5N using the formula:
L' = L + x
L' = L + (F / k)
L' = 35cm + (0.5N / 0.8333 N/cm)
L' = 35cm + 0.6cm
L' = 35.6cm
Therefore, the new length of the cord will be 35.6cm when a force of 0.5N is applied.