Question

two identical springs, each with a spring force constant k, are attached end to end. If a weight is hung from a single spring, it stretches the spring by a distance d. When this same mass is hung from the end of the two springs, which, again, are connected end-to-end, the total stretch of these springs is

Answer 1

Answer:

Δx = 2*d

Explanation:

• According to Hooke's Law, in order to the mass be in equilibrium, when attached to one spring, no net force must act on it, so the algebraic sum of the elastic force and gravity must be zero, as follows:

        $k*\Delta x = m*g (1)$

• If we hang the mass from the end of the two springs attached end to end, in order to be in equilibrium, the total elastic force must be equal to gravity, as we have already said.
• We can express this elastic force, as the product of a Keff times the distance stretched by the two springs combined, as follows:

       $F = k_{eff} * \Delta x_{eff} = m*g (2)$

• Due to F is a tension, it will be the same at any point of the chain of springs, so we can write the following expression, for the distance stretched by any of the springs:

       $\Delta x_{1} = \frac{F}{\ k_{1} } (3)$

• The total distance stretched will be the sum of the distances stretched by any spring individually:

       $\Delta x_{} = \frac{F}{\ k_{1} } + \frac{F}{\ k_{2} } (4)$

• Replacing (4) in (2) and rearranging, we have:

       $\frac{F}{k_{eff} } = \frac{F}{\ k_{1} } + \frac{F}{\ k_{2} } (5)$

      Since k₁ = k₂ = k, we can find keff, as follows:

      $k_{eff} = \frac{k^{2} }{2*k} = \frac{k}{2} (6)$

• Replacing (6) in (2), and making (2) equal to (1) we finally get:

     $F = \frac{k}{2} * \Delta x_{eff} = k*\Delta x = m*g (7)$

• Solving for Δxeff:

       $\Delta x_{eff} = \frac{2*k*\Delta x}{k} = 2* \Delta x = 2*d (8)$