Physics: 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 force of 0.5N,what will be the length of the cord when the stretching force is 2.5N?
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If it has passed its elastic limit, it cannot be predicted.
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Cannot be determined from the given data
An elastic cord can be stretched to its elastic limit by a load of 2n if a 35cm of length 0.5n what will be the length of the cord when the stretching force is 2.5n
To solve this problem, we need to use Hooke's Law, which states that the force required to stretch or compress a spring or elastic material is directly proportional to the amount it is stretched or compressed, assuming the limit of proportionality is not exceeded.
Hooke's Law equation is given by:
F = k * δL
Where:
- F is the force applied
- k is the spring constant (a measure of stiffness), which is specific to the material and geometry of the elastic cord
- δL is the change in length of the cord
In this case, we know that the elastic limit is reached when a load of 2N is applied. We are given that a 35cm length of the cord is extended by 0.6cm under a force of 0.5N.
First, we need to find the spring constant, k. We can use the initial extension to calculate it.
k = F / δL
k = 0.5N / 0.6cm
Now, we can use Hooke's Law to find the length of the cord when the stretching force is 2.5N.
F = 2.5N
k = 0.5N / 0.6cm (as calculated earlier)
F = k * δL
2.5N = (0.5N / 0.6cm) * δL
Solving for δL:
δL = (2.5N * 0.6cm) / 0.5N
δL = 3cm
The length of the cord when the stretching force is 2.5N will be the initial length (35cm) plus the change in length (δL):
Length of the cord = 35cm + 3cm = 38cm
Therefore, the length of the cord when the stretching force is 2.5N will be 38cm.