an eiastic 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 2.5cm.
force is direcdtly proportional to lenght
f2/f1=x2/x1 where x is stretch
F2=.5*2.5/.6
F2=about 2 N
To find the length of the cord when the stretching force is 2.5N, we can start by understanding the relationship between the force applied to the cord and the resulting length of the cord. This relationship is described by Hooke's Law, which states that the force applied to a spring or elastic material is directly proportional to the extension or deformation caused.
Hooke's Law can be written as:
F = k * x
where F is the force applied (in N), k is the spring constant or elastic constant (in N/m), and x is the extension or deformation (in m).
In this case, we know that the elastic limit of the cord is reached when a load of 2N is applied. We are given that a 35cm length of the cord is extended by 0.6cm when a force of 0.5N is applied. We can use this information to find the value of the spring constant (k).
First, let's convert the lengths into meters:
35 cm = 0.35 m
0.6 cm = 0.006 m
Now, we can use Hooke's Law to find the value of k:
0.5N = k * 0.006m
Divide both sides of the equation by 0.006m to isolate k:
k = 0.5N / 0.006m
k = 83.33 N/m
Now that we have the value of the spring constant (k), we can use it to find the length of the cord when a stretching force of 2.5N is applied. Let's call this length "L".
Rearrange Hooke's Law to solve for x:
x = F / k
Substitute the given values:
x = 2.5N / 83.33 N/m
Simplify:
x = 0.03 m
Finally, we can find the length of the cord:
L = Original length + Extension
Original length = 0.35 m
Extension = 0.03 m
L = 0.35 m + 0.03 m
L = 0.38 m
Therefore, the length of the cord when a stretching force of 2.5N is applied will be 0.38 meters.