Question

a spring of negligible mass and length ( l_0 ) hangs from a fixed point. when a mass ( m ) is attached to the free end of the spring, the length of the spring increases to ( l ). the tension in the spring is equal to ( k delta x ), where ( k ) is a constant and ( delta x ) is the extension of the spring. what is ( k )?
a. ( \frac{mg}{l_0} )
b. ( \frac{mg}{l} )
c. ( \frac{mg}{l - l_0} )
d. ( \frac{mg}{l_0 - l} )
∘ a
∘ b
∘ c
∘ d

Explanation:

Step1: Determine the extension of the spring

The extension \(\Delta x\) of the spring is the difference between the final length \(l\) and the original length \(l_0\), so \(\Delta x = l - l_0\).

Step2: Analyze the forces acting on the mass

When the mass \(m\) is in equilibrium, the tension in the spring \(T\) is equal to the weight of the mass \(mg\) (where \(g\) is the acceleration due to gravity). We know that the tension \(T = k\Delta x\).

Step3: Solve for \(k\)

Since \(T = mg\) and \(T = k\Delta x\), we can substitute \(\Delta x = l - l_0\) into \(T = k\Delta x\) to get \(mg = k(l - l_0)\). Then, solving for \(k\) gives \(k=\frac{mg}{l - l_0}\).

Answer:

C. \(\frac{mg}{l - l_0}\)