Question

The ends of spring are attached to blocks of mass 3kg and 2kg. The 3kg block rests on a horizontal surface and the 2kg block which is vertically above it is in equilibrium producing a compression of 1cm of the spring. The 2kg mass must be compressed further by at least _______, so that when it is released, the 3 kg block may be lifted off the ground.

Answer 1

2.5 cm

Answer 2

The problem involves two main stages:

1. Initial Equilibrium: The 2 kg block is in equilibrium, producing a compression of 1 cm (0.01 m) of the spring.
2. Lift-off Condition: The 2 kg block is further compressed and then released. We need to find the minimum additional compression required for the 3 kg block to lift off the ground.

Let $m_1 = 3 \text{ kg}$ and $m_2 = 2 \text{ kg}$. Let $g = 10 \text{ m/s}^2$ (standard approximation, or 9.8 m/s^2, which yields the same result for 'y' as shown in thought process).

Step 1: Calculate the spring constant (k).

In the initial equilibrium state, the spring force balances the weight of the 2 kg block. Let $x_1 = 1 \text{ cm} = 0.01 \text{ m}$ be the initial compression. The spring force $F_s = k x_1$. For equilibrium of the 2 kg block: $k x_1 = m_2 g$ $k (0.01 \text{ m}) = (2 \text{ kg})(10 \text{ m/s}^2)$ $0.01 k = 20$ $k = \frac{20}{0.01} = 2000 \text{ N/m}$

Step 2: Determine the spring extension required to lift the 3 kg block.

The 3 kg block will lift off the ground when the upward force exerted by the spring on it equals its weight. Let $x_{extension}$ be the extension of the spring from its natural length required for lift-off. Upward spring force $F_{lift} = k x_{extension}$. Weight of 3 kg block $W_1 = m_1 g$. For lift-off: $F_{lift} = W_1$ $k x_{extension} = m_1 g$ $(2000 \text{ N/m}) x_{extension} = (3 \text{ kg})(10 \text{ m/s}^2)$ $2000 x_{extension} = 30$ $x_{extension} = \frac{30}{2000} = 0.015 \text{ m} = 1.5 \text{ cm}$

Step 3: Apply Conservation of Energy.

Let 'y' be the additional compression of the 2 kg mass. The total initial compression from the natural length is $X_{initial} = x_1 + y = (0.01 + y)$ meters. The 2 kg block is released from rest from this position. For the 3 kg block to be lifted, the 2 kg block must reach a position where the spring is extended by $x_{extension} = 0.015 \text{ m}$. We consider the minimum condition, where the 2 kg block momentarily comes to rest at the lift-off position (i.e., its kinetic energy is zero at this point).

Let's set the natural length position of the spring as the reference for gravitational potential energy ($h=0$). Initial state (2 kg block compressed by $X_{initial}$):

• Kinetic Energy $K_i = 0$ (released from rest)
• Gravitational Potential Energy $U_{g,i} = m_2 g (-X_{initial})$ (below natural length)
• Spring Potential Energy $U_{s,i} = \frac{1}{2} k X_{initial}^2$

Final state (2 kg block at lift-off position, spring extended by $x_{extension}$):

• Kinetic Energy $K_f = 0$ (minimum condition)
• Gravitational Potential Energy $U_{g,f} = m_2 g (x_{extension})$ (above natural length)
• Spring Potential Energy $U_{s,f} = \frac{1}{2} k x_{extension}^2$

By conservation of mechanical energy ($K_i + U_{g,i} + U_{s,i} = K_f + U_{g,f} + U_{s,f}$): $0 + m_2 g (-X_{initial}) + \frac{1}{2} k X_{initial}^2 = 0 + m_2 g (x_{extension}) + \frac{1}{2} k x_{extension}^2$ $\frac{1}{2} k X_{initial}^2 - m_2 g X_{initial} = \frac{1}{2} k x_{extension}^2 + m_2 g x_{extension}$

Substitute the known values: $k = 2000 \text{ N/m}$, $m_2 = 2 \text{ kg}$, $g = 10 \text{ m/s}^2$, $x_{extension} = 0.015 \text{ m}$. $\frac{1}{2} (2000) X_{initial}^2 - (2)(10) X_{initial} = \frac{1}{2} (2000) (0.015)^2 + (2)(10) (0.015)$ $1000 X_{initial}^2 - 20 X_{initial} = 1000 (0.000225) + 20 (0.015)$ $1000 X_{initial}^2 - 20 X_{initial} = 0.225 + 0.3$ $1000 X_{initial}^2 - 20 X_{initial} = 0.525$

Rearrange into a quadratic equation: $1000 X_{initial}^2 - 20 X_{initial} - 0.525 = 0$

Using the quadratic formula $X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $a = 1000$, $b = -20$, $c = -0.525$ $X_{initial} = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1000)(-0.525)}}{2(1000)}$ $X_{initial} = \frac{20 \pm \sqrt{400 + 2100}}{2000}$ $X_{initial} = \frac{20 \pm \sqrt{2500}}{2000}$ $X_{initial} = \frac{20 \pm 50}{2000}$

Since $X_{initial}$ represents a compression, it must be positive. $X_{initial} = \frac{20 + 50}{2000} = \frac{70}{2000} = 0.035 \text{ m}$

This $X_{initial}$ is the total compression from the natural length. We know $X_{initial} = x_1 + y$. $0.035 \text{ m} = 0.01 \text{ m} + y$ $y = 0.035 \text{ m} - 0.01 \text{ m}$ $y = 0.025 \text{ m}$

Converting to centimeters: $y = 2.5 \text{ cm}$

The 2 kg mass must be compressed further by at least 2.5 cm.

The final answer is $\boxed{2.5 \text{ cm}}$.

Explanation of the solution:

1. Determine spring constant (k): Use the initial equilibrium condition where the 2kg mass is balanced by the spring force, with 1cm compression. $k \times (0.01 \text{ m}) = 2 \text{ kg} \times g$.
2. Determine extension for lift-off: Calculate the spring extension required for the spring force to equal the weight of the 3kg block. $k \times x_{extension} = 3 \text{ kg} \times g$.
3. Apply energy conservation: Set up the energy conservation equation for the 2kg block and the spring. The initial state is the 2kg block compressed by $(0.01 + y)$ from natural length, released from rest. The final state is the 2kg block at the position where the spring is extended by $x_{extension}$, with minimum kinetic energy (zero). Solve the resulting quadratic equation for the total initial compression $(0.01+y)$, then find 'y'.