Physics Solution: Block, Plank, and Spring System
1. Problem Statement Analysis
A block ($m=1.0 \text{ kg}$) is given an initial velocity $v_0$ on a plank ($M=15 \text{ kg}$). It compresses a spring ($k=500 \text{ N/m}$) and stops relative to the plank without reversing. We need to find $v_0$.
Friction coefficient $\mu = 0.5$.
2. Concept: Stopping without Reversing
The condition “stops on the plank before reversing” implies that at the point of maximum compression ($x_{max}$), the system comes to a state of relative rest ($v_{rel}=0$) and stays there. This means the restoring force of the spring is balanced by the limiting static friction.
At equilibrium position $x_{max}$:
$$F_{spring} = f_{friction}$$ $$k x_{max} = \mu m g$$From this, we can find the maximum compression:
$$x_{max} = \frac{\mu m g}{k}$$Substituting values ($g=10 \text{ m/s}^2$):
$$x_{max} = \frac{0.5 \times 1.0 \times 10}{500} = \frac{5}{500} = 0.01 \text{ m}$$3. Work-Energy Theorem (Reduced Mass Frame)
Using the conservation of energy in the frame of reference of the Center of Mass (or using reduced mass concept), the initial relative kinetic energy is converted into the potential energy of the spring and the work done against friction.
Initial Kinetic Energy (Relative):
$$KE_{rel} = \frac{1}{2} \mu_{red} v_{rel}^2 = \frac{1}{2} \left( \frac{mM}{m+M} \right) v_0^2$$Energy Balance Equation:
$$KE_{rel} = PE_{spring} + W_{friction}$$ $$\frac{1}{2} \left( \frac{mM}{m+M} \right) v_0^2 = \frac{1}{2} k x_{max}^2 + \mu m g x_{max}$$Substitute $x_{max} = \frac{\mu mg}{k}$:
$$\frac{1}{2} \left( \frac{mM}{m+M} \right) v_0^2 = \frac{1}{2} k \left( \frac{\mu mg}{k} \right)^2 + \mu mg \left( \frac{\mu mg}{k} \right)$$ $$\frac{1}{2} \left( \frac{mM}{m+M} \right) v_0^2 = \frac{(\mu mg)^2}{2k} + \frac{(\mu mg)^2}{k} = \frac{3(\mu mg)^2}{2k}$$4. Solving for Velocity $v_0$
Canceling $\frac{1}{2}$ from both sides:
$$\left( \frac{mM}{m+M} \right) v_0^2 = \frac{3(\mu mg)^2}{k}$$ $$v_0^2 = \frac{3(\mu mg)^2 (m+M)}{k m M}$$Plugging in the numbers:
- $\mu m g = 5 \text{ N}$
- $m + M = 16 \text{ kg}$
- $mM = 15 \text{ kg}^2$
- $k = 500 \text{ N/m}$