Two spring block system lies on smooth horizontal surface the free end of spring is beibg pulled right with speed 2m/s.at t=0 the spring is unstretched. and the block has speed 1 m/s. to left the max. elongation is ? (k=100N/cm)

To solve the problem of the two-spring block system, we need to analyze the forces and energy involved when the spring is stretched. The key here is to understand how the kinetic energy of the block converts into potential energy stored in the spring when it reaches maximum elongation.

Understanding the System

We have a spring with a spring constant \( k = 100 \, \text{N/cm} \), which we can convert to standard units: \( k = 10000 \, \text{N/m} \) (since 1 N/cm = 100 N/m). The block is initially moving to the left with a speed of \( 1 \, \text{m/s} \), while the free end of the spring is being pulled to the right at \( 2 \, \text{m/s} \). This means that the relative speed between the block and the end of the spring is \( 2 + 1 = 3 \, \text{m/s} \).

Energy Considerations

At maximum elongation, the kinetic energy of the block will be completely converted into the potential energy of the spring. The kinetic energy (\( KE \)) of the block can be calculated using the formula:

• KE = \frac{1}{2} mv^2

Where \( m \) is the mass of the block and \( v \) is its velocity. However, we don't have the mass of the block, so we will keep it as \( m \) for now.

Potential Energy of the Spring

The potential energy (\( PE \)) stored in the spring when it is elongated by a distance \( x \) is given by:

• PE = \frac{1}{2} k x^2

Setting Up the Equation

At maximum elongation, all the kinetic energy of the block will equal the potential energy of the spring:

• \frac{1}{2} mv^2 = \frac{1}{2} k x^2

Substituting the values we know:

• \frac{1}{2} m (3)^2 = \frac{1}{2} (10000) x^2

This simplifies to:

• 9m = 10000 x^2

Solving for Maximum Elongation

Rearranging the equation gives us:

• x^2 = \frac{9m}{10000}

Taking the square root of both sides, we find:

• x = \sqrt{\frac{9m}{10000}} = \frac{3\sqrt{m}}{100}

Conclusion

The maximum elongation \( x \) of the spring depends on the mass of the block. If you know the mass, you can substitute it into the equation to find the exact elongation. For example, if the mass of the block is \( 1 \, \text{kg} \), then:

• x = \frac{3\sqrt{1}}{100} = 0.03 \, \text{m} = 3 \, \text{cm}

This means that the maximum elongation of the spring will be \( 3 \, \text{cm} \) if the block has a mass of \( 1 \, \text{kg} \). Adjust the mass accordingly to find the elongation for different values. This approach illustrates the conservation of energy principle in a mechanical system.

Approved