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

Answer 1

First, let's calculate the spring constant of the elastic cord using the initial load and stretch:

k = F/x
k = 2N / 0.6cm
k = 3.33 N/cm

Now, we can calculate the change in length of the cord when the stretching force is 0.5N:

Δx = F / k
Δx = 0.5N / 3.33N/cm
Δx = 0.15cm

The total length of the cord when the stretching force is 0.5N will be:

35cm + 0.15cm = 35.15cm

Now, we can calculate the change in length of the cord when the stretching force is 2.5N:

Δx = F / k
Δx = 2.5N / 3.33N/cm
Δx = 0.75cm

The total length of the cord when the stretching force is 2.5N will be:

35.15cm + 0.75cm = 35.9cm

Therefore, the length of the cord when the stretching force is 2.5N will be 35.9cm.