A spring of natural length 1.5m is extended 0.055cm by force of 0.8N, what will it length be when the applied force is 3.2N
area you sure you mean .055 cm and not .055 m ?
F = k x
.8 = k *.00055
x + 1.5 = F/k + 1.5 = 3.2 *.00055/.8 + 1.5
0n
To answer this question, we can use Hooke's Law, which states that the force needed to extend or compress a spring is directly proportional to the displacement.
The formula for Hooke's Law is:
F = k * x
Where:
F is the force applied to the spring
k is the spring constant
x is the displacement of the spring
In order to find the length of the spring when the applied force is 3.2N, we first need to determine the spring constant, k.
Using the given information:
Natural length, L0 = 1.5m
Extension, x = 0.055cm = 0.055 * 0.01 = 0.00055m
Force applied, F = 0.8N
We can calculate the spring constant using the formula:
k = F / x
Substituting the values we have:
k = 0.8N / 0.00055m
k ≈ 1454.545 N/m (rounded to 3 decimal places)
Now that we have the spring constant, we can determine the length of the spring when the applied force is 3.2N.
Using the formula:
F = k * x
Rearranging the formula to isolate x:
x = F / k
Substituting the values we have:
x = 3.2N / 1454.545 N/m
x ≈ 0.0022m (rounded to 4 decimal places)
Finally, we find the length of the spring by adding the displacement to the natural length:
Length = L0 + x
Substituting the values we have:
Length = 1.5m + 0.0022m
Length ≈ 1.5022m (rounded to 4 decimal places)
Therefore, the length of the spring will be approximately 1.5022 meters when the applied force is 3.2 Newtons.