A block of mass m moving at a speed 'v' compresses as spring through a distance 'x' before its speed is halved. Find the spring constant of the spring.

Explanation

To find the spring constant \( k \) of the spring when a block of mass \( m \) moving at speed \( v \) compresses the spring through a distance \( x \) before its speed is halved, we can use the principle of conservation of energy. Here’s the step-by-step solution:

### Step 1: Write the initial kinetic energy
The initial kinetic energy \( KE_i \) of the block when it is moving at speed \( v \) is given by:
\[ KE_i = \frac{1}{2} m v^2 \]

### Step 2: Write the final kinetic energy
When the block compresses the spring by distance \( x \), its speed is halved. Therefore, the final speed \( v_b \) is:
\[ v_b = \frac{v}{2} \]
The final kinetic energy \( KE_f \) is then:
\[ KE_f = \frac{1}{2} m v_b^2 = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{2} m \frac{v^2}{4} = \frac{1}{8} m v^2 \]

### Step 3: Write the potential energy stored in the spring
The potential energy \( PE \) stored in the spring when it is compressed by distance \( x \) is given by:
\[ PE = \frac{1}{2} k x^2 \]

### Step 4: Apply the conservation of energy principle
According to the conservation of energy, the initial kinetic energy is equal to the sum of the final kinetic energy and the potential energy stored in the spring:
\[ KE_i = KE_f + PE \]
Substituting the expressions we derived:
\[ \frac{1}{2} m v^2 = \frac{1}{8} m v^2 + \frac{1}{2} k x^2 \]

### Step 5: Simplify the equation
First, let's eliminate \( m \) from the equation (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 = \frac{1}{8} v^2 + \frac{1}{2} k x^2 \]
Now, multiply the entire equation by 8 to eliminate the fractions:
\[ 4 v^2 = v^2 + 4 k x^2 \]

### Step 6: Rearrange to solve for \( k \)
Subtract \( v^2 \) from both sides:
\[ 4 v^2 - v^2 = 4 k x^2 \]
\[ 3 v^2 = 4 k x^2 \]
Now, isolate \( k \):
\[ k = \frac{3 v^2}{4 x^2} \]

### Final Answer
The spring constant \( k \) is given by:
\[ k = \frac{3 m v^2}{4 x^2} \]