Question

Block of mass m is released from the position A when spring was in its natural length. A constant force $F_o$ also starts acting on mass at time of release then maximum elongation in the spring will be

Answer 1

$\frac{2(mg + F_o)}{K}$

Answer 2

The problem asks for the maximum elongation in a spring when a block of mass 'm' is released from rest. Initially, the spring is at its natural length. A constant force $F_o$ also acts on the mass downwards.

1. Identify the initial and final states:

• Initial state (A): The block is at rest ($v_i = 0$). The spring is at its natural length (spring potential energy $U_{s,i} = 0$). Let's set the gravitational potential energy reference at position A ($U_{g,i} = 0$).

• Final state (B): The block reaches its maximum elongation, meaning it momentarily comes to rest ($v_f = 0$). Let the maximum elongation be 'x'.

2. Apply the Work-Energy Theorem:

The Work-Energy Theorem states that the net work done on an object equals the change in its kinetic energy: $W_{net} = \Delta KE = KE_f - KE_i$

Since the block starts from rest and momentarily stops at maximum elongation, both initial and final kinetic energies are zero: $KE_i = 0$ $KE_f = 0$ Therefore, $\Delta KE = 0$. So, $W_{net} = 0$.

3. Calculate the work done by each force:

The forces acting on the block are:

• Gravity (mg): Acts downwards. As the block moves downwards by a distance 'x', the work done by gravity is $W_g = mgx$.

• Constant force ($F_o$): Acts downwards. As the block moves downwards by a distance 'x', the work done by $F_o$ is $W_{F_o} = F_o x$.

• Spring force (kx): Acts upwards (opposite to the displacement) as the spring elongates downwards. The spring force is a variable force. The work done by the spring force is $W_s = -\frac{1}{2}kx^2$.

4. Set up the Work-Energy equation:

Sum of work done by all forces equals zero: $W_g + W_{F_o} + W_s = 0$ $mgx + F_o x - \frac{1}{2}kx^2 = 0$

5. Solve for the maximum elongation (x):

Factor out 'x' from the equation: $x \left( mg + F_o - \frac{1}{2}kx \right) = 0$

This equation gives two possible solutions:

• $x = 0$: This corresponds to the initial position where the spring is at its natural length.

• $mg + F_o - \frac{1}{2}kx = 0$: This corresponds to the position of maximum elongation.

From the second solution: $\frac{1}{2}kx = mg + F_o$ $x = \frac{2(mg + F_o)}{K}$

This is the maximum elongation in the spring.

The options provided in the question (e.g., -$\mu$mgH, -mgH, -$\mu$mgL) are expressions for work or energy, not for elongation. They seem to belong to a different question. Therefore, the answer is the derived expression for maximum elongation.

The final answer is $\boxed{\frac{2(mg + F_o)}{K}}$