An elastic cord can be stretched to it's elastic limit by a load of 2N . If a 35cm length of the cord is extended 0.6cm by a force of 0.5 n what will be the length of the cord when the stretching force is 2.5n
We can use the concept of proportionality of deformation to determine the length of the cord when the stretching force is 2.5N.
Let's assume that the initial length of the cord is L and the extension produced by a force of 0.5N is x. We can then define the elastic constant of the cord as k = (2N)/L, since this force value represents the elastic limit of the cord.
Using this information, we can write:
x/L = F/k
where F is the stretching force and k is the elastic constant.
We can rearrange this equation to solve for L:
L = x/(F/k) = (xk)/F
Substituting the given values, we get:
L = (0.6 cm)((2 N)/L)/(0.5 N)
L = 2.4 cm
Therefore, the initial length of the cord is 35 cm and it extends to 37.4 cm when stretched by a force of 2.5N.
To determine the length of the cord when the stretching force is 2.5N, we can use the concept of Hooke's Law which states that the extension of an elastic material is directly proportional to the force applied.
First, let's calculate the constant of proportionality (k) using the given information that the elastic cord can be stretched to its elastic limit by a load of 2N when a 35cm length of the cord is extended by 0.6cm:
k = (Extension) / (Force)
k = 0.6 cm / 2 N
k = 0.3 cm/N
Now, we can use this constant to find the extension of the cord when a force of 2.5N is applied:
Extension = k * Force
Extension = 0.3 cm/N * 2.5 N
Extension = 0.75 cm
Finally, we can calculate the final length of the cord by adding the extension to the original length:
Final Length = Original Length + Extension
Final Length = 35 cm + 0.75 cm
Final Length = 35.75 cm
Therefore, the length of the cord when the stretching force is 2.5N will be 35.75 cm.