An elastic cord can be stretched to its elastic limit by a load of 2N. If a 35cm length of the cord is extende 0.6cm by a force of the cord when the stretching force is 2.5N
solve the equation; an elastic cord can be stretched to its elastic limit by a load of 2N. if 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. a) 350.8cm. b) 352.8cm. c) 353.0cm. d) 356cm.
Cryptic indeed
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Cannot be determined by the given data
To solve this problem, we can use Hooke's Law, which states that the force needed to extend or compress a spring (or an elastic cord) is directly proportional to the extension or compression of the spring, as long as the elastic limit is not exceeded.
Hooke's Law can be expressed as:
F = k * x
Where:
F is the force applied to the cord
k is the spring constant (a measure of its stiffness)
x is the extension or compression of the cord
We are given that the elastic limit of the cord is reached at a load of 2N. This means that when the force applied to the cord is greater than 2N, it will no longer behave elastically and may be permanently deformed.
We are also given that a 35cm length of the cord is extended by 0.6cm when a force of 2.5N is applied to it. We can use this information to calculate the spring constant, k.
First, let's convert the given lengths to meters:
Length of the cord, L = 35cm = 0.35m
Extension of the cord, Δx = 0.6cm = 0.006m
Next, we can rearrange Hooke's Law to solve for the spring constant, k:
k = F / x
Using the values given in the problem, we can substitute them into the formula:
k = 2.5N / 0.006m
Now we can calculate the spring constant:
k = 416.67 N/m
Finally, we are asked to find the force required to stretch the cord when the stretching force is 2.5N. We can use Hooke's Law again:
F = k * x
Substituting the values we have:
F = 416.67 N/m * 0.006m
F = 2.5N
Therefore, the force required to stretch the cord when the stretching force is 2.5N is 2.5N.