Question

A block of mass \(5 kg\) moves along the \(x-\)direction subject to the force \(F = (−20x + 10) N,\) with the value of  \(x \) in metre. At time \(t = 0 s,\) it is at rest at position \(x = 1 m\). The position and momentum of the block at \(t = (\pi/4)\) s are

• A. \(-0.5m,5kg \ \frac{m}{s}\)
• B. \(0.5m,0kg \ \frac{m}{s}\)
• C. \(0.5m,0kg \ \frac{m}{s}\)
• D. \(0.5m,0kg \ \frac{m}{s}\)

Answer 1

The Correct Option is C

Motion of the Block and Simple Harmonic Motion (SHM) 

1. Force Acting on the Block

The force acting on the block is given by:

\[ F = -20x + 10. \]

2. Acceleration Using Newton’s Second Law

Using Newton’s second law, \( F = ma \), where \( m = 5 \) kg:

\[ a = \frac{F}{m} = \frac{-20x + 10}{5} = -4x + 2. \]

3. Equilibrium Position

At equilibrium, \( a = 0 \):

\[ -4x + 2 = 0 \Rightarrow x = 0.5 \text{ m}. \]

This is the **mean position (M.P.)** of the block.

4. Equation of Motion

For small oscillations about the equilibrium position:

\[ a = -4(x - 0.5). \]

Comparing with the standard SHM equation \( a = -\omega^2 x \), we get:

\[ \omega^2 = 4 \Rightarrow \omega = 2 \text{ rad/s}. \]

The general equation of motion is:

\[ x = 0.5 + A \cos(\omega t). \]

5. Finding Amplitude

At \( t = 0 \), \( x = 1 \) m:

\[ 1 = 0.5 + A \Rightarrow A = 0.5. \]

Thus, the position function becomes:

\[ x = 0.5 + 0.5 \cos(2t). \]

6. Position at \( t = \frac{\pi}{4} \)

\[ x = 0.5 + 0.5 \cos\left(2 \times \frac{\pi}{4}\right) = 0.5 + 0.5 \cdot 0 = 0.5 \text{ m}. \]

7. Velocity Function

The velocity is given by:

\[ v = \frac{dx}{dt} = -0.5 \cdot 2 \sin(2t) = -\sin(2t). \]

At \( t = \frac{\pi}{4} \):

\[ v = -\sin\left(2 \times \frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{2}\right) = -1 \text{ m/s}. \]

8. Momentum Calculation

Momentum is given by:

\[ p = m \cdot v = 5 \times (-1) = -5 \text{ kg m/s}. \]

Final Answer:

• Position: 0.5 m
• Momentum: -5 kg m/s