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?
Huh???
perhaps infinite
There is a good chance it broke. You can not use Hooke's law beyond the elastic limit.
To solve this problem, we can use Hooke's Law, which states that the force applied to an elastic object is directly proportional to the displacement of the object.
Hooke's Law can be written as F = kx, where F is the force applied, k is the spring constant (a property of the elastic cord), and x is the displacement of the cord.
To find the constant k, we can use the given information. It states that the elastic limit of the cord is reached when a load of 2N is applied. Since the displacement under this load is not given, we can assume it to be x_1.
Therefore, we have:
2N = k * x_1
Next, we can use the given information that a 35cm length of the cord is extended by 0.6cm with a force of 0.5N. Let's assume this displacement to be x_2.
Therefore, we have:
0.5N = k * x_2
To find the value of k, we can solve the first equation for k:
k = 2N / x_1
Now, we can substitute the value of k into the second equation to find x_2:
0.5N = (2N / x_1) * x_2
Simplifying the equation by canceling N and rearranging:
x_1 = 4 * x_2
This equation gives us the ratio of the displacements for a given elastic cord.
Now, we can use this ratio to find the displacement x_3 when a force of 2.5N is applied.
x_3 = (2.5N / 0.5N) * x_2
Simplifying:
x_3 = 5 * x_2
Finally, we have the ratio of displacements between the load of 0.5N and 2.5N. We can use this ratio along with the given displacement of 0.6cm to find the final length of the cord.
length_3 = 35cm + ((x_3 - x_2) * 100)
Substituting the values:
length_3 = 35cm + ((5 * 0.6cm - 0.6cm) * 100)
Simplifying:
length_3 = 35cm + (3 * 100)
length_3 = 35cm + 300cm
length_3 = 335cm
Therefore, the length of the cord when the stretching force is 2.5N will be 335cm.