A coil spring is extended by 4.4 cm when it supports a 1.00- N weight. If an identical spring is joined to the end of the first one, what is the extension of the combined spring when it supports the same 1.00- N weight? Assume the spring masses to be negligible
To find the extension of the combined spring when it supports the same 1.00 N weight, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to its extension.
First, let's find the spring constant (k) of each individual spring.
Hooke's Law equation: F = k * x
Where:
F = force applied (1.00 N)
k = spring constant
x = extension of the spring (4.4 cm or 0.044 m)
Rearrange the equation to solve for k:
k = F / x
k = 1.00 N / 0.044 m
k ≈ 22.73 N/m
Now, we have the spring constant (k) for each spring.
To find the extension of the combined spring, we can model this situation as two springs in series. When springs are in series, their effective spring constant (k_eff) is given by:
1 / k_eff = 1 / k₁ + 1 / k₂
Where:
k_eff = effective spring constant of the combined spring
k₁ = spring constant of the first spring
k₂ = spring constant of the second spring
Substituting in the values:
1 / k_eff = 1 / 22.73 N/m + 1 / 22.73 N/m
Simplifying the equation, we get:
1 / k_eff = 2 / 22.73 N/m
1 / k_eff = 0.088
Solving for k_eff:
k_eff = 1 / 0.088
k_eff ≈ 11.36 N/m
Now, we have the effective spring constant (k_eff) of the combined spring.
Finally, to find the extension of the combined spring when it supports the same 1.00 N weight, we can use Hooke's Law again:
F = k_eff * x_combined
Rearranging the equation to solve for x_combined:
x_combined = F / k_eff
x_combined = 1.00 N / 11.36 N/m
x_combined ≈ 0.088 m or 8.8 cm
Therefore, the extension of the combined spring when it supports the same 1.00 N weight is approximately 0.088 m or 8.8 cm.