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 will be the length of the cord when the stretching force is 2.5N ?
We can use Hooke's law to solve this problem:
F = kx
where F is the stretching force, x is the extension, and k is the spring constant.
We know that the elastic limit of the cord is 2N, so we can calculate the spring constant as:
k = F/x = 2N/0.6cm = 3.33 N/cm
Now, we can use this spring constant to calculate the length of the cord when the stretching force is 2.5N:
x = F/k = 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:
L = 35cm + 0.75cm = 35.75cm
To determine the length of the cord when the stretching force is 2.5N, we can first find the spring constant (k) using Hooke's Law for the given values of force (0.5N) and extension (0.6cm).
Hooke's Law states: F = k * x
Where F is the force, k is the spring constant, and x is the extension.
1. Find the spring constant (k).
0.5N = k * 0.6cm
k = 0.5N / 0.6cm
k ≈ 0.8333 N/cm
2. Now we can find the extension (x) when the force is 2.5N.
2.5N = 0.8333 N/cm * x
x = 2.5N / 0.8333 N/cm
x ≈ 3cm
3. Finally, we can calculate the length of the cord (L) when the stretching force is 2.5N.
L = original length + extension
L = 35cm + 3cm
L = 38cm
So, the length of the cord when the stretching force is 2.5N will be approximately 38cm.