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

The potential energy lost by the block is converted into the elastic potential energy of the spring. Using the conservation of mechanical energy: \[ mgh = \frac{1}{2} k x^2 \] Where: - \( m \) is the mass of the block, - \( g = 9.81 \, \text{m/s}^2 \), - \( h = 0.6 \, \text{m} \), - \( k = 2400 \, \text{N/m} \), - \( x = 0.25 - 0.10 = 0.15 \, \text{m} \) (the compression of the spring). Substitute the values into the equation: \[ m \times 9.81 \times 0.6 = \frac{1}{2} \times 2400 \times (0.15)^2 \] Solving for \( m \): \[ m = \frac{2400 \times (0.15)^2}{2 \times 9.81 \times 0.6} \approx 55.51 \, \text{kg} \]