An elastic cord can be stretched to it elastic limit by a load of 4N. If a 40cm length of the cord is extended to 0.8cm by a force of 0.5N, What will be the length of the cord when stretcher point is 2.5N.

Answer 1

To find the length of the cord when stretched by a force of 2.5N, we need to use the concept of Hooke's Law.

Hooke's Law states that the force required to stretch or compress a spring by a distance x is directly proportional to the distance x and the spring constant k.

In this case, the elastic cord obeys Hooke's Law. We are given that the cord can be stretched to its elastic limit by a load of 4N, giving us the force and displacement information as follows:

Force (F) = 4N
Initial length (L0) = 40cm = 0.40m
Final length (L) = 0.8cm = 0.008m

Using Hooke's Law equation:
F = k * x

where F is the force, k is the spring constant, and x is the displacement.

We need to find the spring constant k by rearranging the formula:

k = F / x

Now we can calculate the spring constant:
k = 4N / 0.008m = 500 N/m

We can now use the spring constant to find the length of the cord when stretched by a force of 2.5N.

Force (F) = 2.5N
k = 500 N/m

Using the rearranged formula:
F = k * x

Rearranging for x:
x = F / k

x = 2.5N / 500 N/m = 0.005m

Finally, we can calculate the final length (L) of the cord when stretched by a force of 2.5N:

L = L0 + x
L = 0.40m + 0.005m = 0.405m

Therefore, the final length of the cord will be 0.405m when stretched by a force of 2.5N.