A spring of natural length 30cm is stretched 35cm by a load of 15N, calculate the extension produced
You just told us it was extended 35cm.
What difference does the force make?
Why have you not reworded the question so it makes sense?
To calculate the extension produced by the load on the spring, we need to use Hooke's Law, which states that the extension of a spring is directly proportional to the force applied to it. Hooke's Law can be expressed as:
F = kx
Where F is the force applied, k is the spring constant, and x is the extension produced.
In this case, we are given the force applied (15N) and the natural length of the spring (30cm). However, we need to convert the measurements to a consistent unit. Let's convert the natural length to meters first:
Natural length = 30cm = 0.3m
Now, let's rearrange Hooke's Law to solve for the extension (x):
x = F / k
Since we don't have the spring constant (k) given, we can calculate it using the formula:
k = F / x
Substituting the given values into the formula:
k = 15N / 0.35m = 42.857 N/m (rounded to three decimal places)
Now we can use the calculated spring constant to find the extension:
x = F / k = 15N / 42.857 N/m = 0.35m (rounded to two decimal places)
Therefore, the extension produced by the load is 0.35m.