Question

A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is 2 m and spring constant is 200 N/m. The block is pushed such that the length of the spring becomes 1 m and then released. At distance $ x $ m ($ x \leq 2 $) from the wall, the speed of the block will be:

• A. \( 10 \left[ 1 - (2 - x)^2 \right] \, \text{m/s} \)
• B. \( 10 \left[ 1 - (2 - x) \right]^{3/2} \, \text{m/s} \)
• C. \( 10 \left[ 1 - (2 - x)^2 \right]^{1/2} \, \text{m/s} \)
• D. \( 10 \left[ 1 - (2 - x)^2 \right]^2 \, \text{m/s} \)

Hint:

In spring-mass systems, use conservation of mechanical energy to relate potential energy and kinetic energy. The speed of the block can be found by equating the total energy at different points in the motion.

Answer 1

The Correct Option is C

The total mechanical energy is the sum of the potential energy stored in the spring and the kinetic energy of the block. The potential energy \( U \) stored in a spring is given by: \( U = \frac{1}{2} k (x_{\text{spring}})^2 \), 
where \( k = 200 \, \text{N/m} \) is the spring constant, and \( x_{\text{spring}} \) is the displacement from the natural length of the spring. The initial displacement of the spring is 1 m (the block is compressed), so the initial potential energy is: 
\( U_{\text{initial}} = \frac{1}{2} \times 200 \times (2 - 1)^2 = 100 \, \text{J} \).

The total mechanical energy of the system is constant and is the sum of the kinetic energy \( K = \frac{1}{2} m v^2 \) and the potential energy stored in the spring at any point during the motion.<br>
The total energy \( E \) is given by: 
\( E = K + U \). 
At any position \( x \), the total energy is: 
\( E = \frac{1}{2} m v^2 + \frac{1}{2} k (2 - x)^2 \).

Since the total mechanical energy is conserved and the initial energy is all potential energy, we have: 
\( 100 = \frac{1}{2} \times 2 \times v^2 + \frac{1}{2} \times 200 \times (2 - x)^2 \). 
Simplifying: 
\( 100 = v^2 + 200 (2 - x)^2 \).

Solving for \( v \): 
\( v^2 = 100 - 200 (2 - x)^2 \), 
\( v = 10 \left[ 1 - (2 - x)^2 \right]^{1/2} \, \text{m/s} \).

Thus, the speed of the block at distance \( x \) from the wall is: 
\( \boxed{10 \left[ 1 - (2 - x)^2 \right]^{1/2}} \, \text{m/s} \).

Therefore, the correct answer is Option (3).