Question

A sphere of mass 25 gram is placed on a vertical spring. It is compressed by 0.2 m using a force 5 N. When the spring is released, the sphere will reach a height of \( g = 10 \, \text{m/s}^2 \)

• A. 6 cm
• B. 8 cm
• C. 10 cm
• D. 2 m

Hint:

Use energy conservation from the fully compressed position to the highest point. The spring potential energy converts entirely to gravitational potential energy (since at the highest point the sphere is momentarily at rest and the spring is relaxed). Do not forget to convert mass to kg.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Mass \(m = 25\ \text{g} = 0.025\ \text{kg}\). Spring compressed by \(x = 0.2\ \text{m}\) using force \(F = 5\ \text{N}\). Spring constant \(k = F/x = 5 / 0.2 = 25\ \text{N/m}\). The sphere is placed on the spring (presumably on top), spring is compressed, then released. The sphere will be projected upward. Find maximum height reached.

Step 2: Key Formula or Approach:
Energy conservation: Elastic potential energy stored in spring = gravitational potential energy gained by sphere (if we neglect air resistance and assume spring’s mass negligible). But careful: When released, the spring expands to natural length, then beyond that the sphere loses contact. Actually the sphere will leave contact at the natural length. The energy stored initially \(\frac{1}{2} k x^2\) converts to kinetic energy at natural length, then that kinetic energy converts to height. Alternatively, from compressed position to maximum height, the spring does work. But easier: Use energy conservation from initial compressed position to the highest point (where velocity = 0). At the highest point, the sphere has gained gravitational PE relative to the initial compressed position. However, the spring does positive work until natural length, then the sphere continues upward under gravity. But the total mechanical energy (spring + gravitational) is conserved. Choose zero gravitational PE at the initial compressed position. Then initial energy = spring PE = \(\frac{1}{2} k x^2\). At final maximum height \(h\) above the initial position, the sphere has gravitational PE = \(m g h\) (since spring is relaxed, no spring PE). So \(\frac{1}{2} k x^2 = m g h\). Thus \(h = \frac{k x^2}{2 m g}\).

Step 3: Detailed Explanation:
Compute: \(k = 25\ \text{N/m}\), \(x = 0.2\ \text{m}\), \(m = 0.025\ \text{kg}\), \(g = 10\ \text{m/s}^2\).
Spring PE = \(0.5 \times 25 \times (0.2)^2 = 0.5 \times 25 \times 0.04 = 0.5 \times 1 = 0.5\ \text{J}\).
Then \(h = \frac{0.5}{m g} = \frac{0.5}{0.025 \times 10} = \frac{0.5}{0.25} = 2\ \text{m}\).
That gives 2 m, which is option (D). But wait the options include 2 m as (D). However earlier I thought 10 cm? Let's double-check. The problem statement says "the sphere will reach a height of \( g = 10 \, \text{m/s}^2 \) 2 m" - the phrasing is awkward. Actually reading: "will reach a height of \( g = 10 \, \text{m/s}^2 \) 2 m" maybe means options are A 6 cm, B 8 cm, C 10 cm, D 2 m. With calculation we got 2 m. So answer should be (D) 2 m. But my quick tip says 10 cm? Let me recalc carefully: \(k = F/x = 5/0.2 = 25\) N/m. \(x=0.2\) m. \(m=0.025\) kg. \(g=10\). Then \(\frac{1}{2} k x^2 = 0.5 * 25 * 0.04 = 0.5 * 1 = 0.5\) J. \(mgh = 0.025 * 10 * h = 0.25 h\). Equate: \(0.25 h = 0.5\) → \(h = 2\) m. Yes. So correct answer is 2 m, option (D). My earlier mistake.

Step 4: Final Answer:
Option (D) 2 m.