Question

A block falls from a table 0.6 m high. It lands on an ideal, mass-less, vertical spring with a force constant of 2.4 kN/m. The spring is initially 25 cm high, but it is compressed to a minimum height of 10 cm before the block is stopped. Find the mass of the block (g = 9.81 m/s\(^2\)):

• A. 55.51 kg
• B. 5.51 kg
• C. 0.51 kg
• D. None

Hint:

The principle of conservation of mechanical energy helps in solving such problems involving potential energy and elastic potential energy.

Answer 1

The Correct Option is A

Step 1: Split the fall into two stages: free fall from the table down to the spring's natural top, a drop of \(0.6 - 0.25 = 0.35\,\text{m}\), then compression of the spring by a further \(0.25-0.10=0.15\,\text{m}\).

Step 2: During the compression stage the block keeps descending, so it also loses an extra \(mg(0.15)\) of potential energy on top of the kinetic energy it already carries from the free fall. All of this combined energy converts into spring energy at maximum compression: \( mg(0.35) + mg(0.15) = \frac{1}{2}kx^2 \), the same as \( mg(0.5) = \frac{1}{2}kx^2 \).

Step 3: Substitute \(k=2400\,\text{N/m}\), \(x=0.15\,\text{m}\), \(g=9.81\,\text{m/s}^2\): \( \frac{1}{2}(2400)(0.15)^2 = 27\,\text{J} \), so \( m(9.81)(0.5)=27 \).

Step 4: Solve: \( m = \dfrac{27}{4.905} \approx 5.51\,\text{kg}\).\[ \boxed{m \approx 5.51 \, \text{kg}} \]

Answer 2

Solve the energy balance symbolically before plugging in any numbers: from \(mgh=\frac{1}{2}kx^2\), the mass is \(m = \dfrac{kx^2}{2gh}\). Checking units first is a good habit here: \(k\) is in N/m, \(x^2\) in \(\text{m}^2\), so \(kx^2\) is in N\(\cdot\)m, i.e. joules; dividing by \(g\) (m/s\(^2\)) and \(h\) (m) leaves units of kilograms, confirming the formula gives a mass as expected.

Now substitute the numbers: spring constant \(k=2400\,\text{N/m}\), compression \(x=0.25-0.10=0.15\,\text{m}\), effective drop \(h=0.6-0.10=0.5\,\text{m}\), and \(g=9.81\,\text{m/s}^2\):\[ m = \frac{(2400)(0.15)^2}{2(9.81)(0.5)} = \frac{54}{9.81} \]

Carrying out the division gives \(m \approx 5.51\,\text{kg}\), consistent with the units check performed above.\[ \boxed{m \approx 5.51\,\text{kg}} \]