Physics 1

## Opposing Force Against Penetration ### Problem Statement A hammer of mass \( M \) falls from a height \( h \) to drive a nail of mass \( m \) into the ground. The hammer makes the nail penetrate a distance \( d \) into the ground. Find the opposing force against the penetration. ### Step-by-Step Solution #### Step 1: Understand the Scenario The hammer falls from a height \( h \), converting its potential energy into kinetic energy just before hitting the nail. Upon impact, the hammer and nail move together (assuming they stick or move as one object after impact), penetrating the ground to a depth \( d \) against an opposing force. #### Step 2: Calculate Velocity of Hammer Before Impact The potential energy of the hammer at height \( h \) is \( Mgh \). Just before impact, all this energy converts into kinetic energy: \[ \frac{1}{2}Mv^2 = Mgh \] Solving for \( v \): \[ v = \sqrt{2gh} \] #### Step 3: Apply Conservation of Momentum (Optional for This Direct Approach) However, since the question focuses on the force opposing penetration and not the details of the collision, we'll directly consider energy and work done during penetration. #### 4: Work Done Against Opposing Force The work done by the hammer and nail system against the opposing force during penetration is: \[ W = F \cdot d \] where \( F \) is the opposing force. #### 5: Energy Consideration The kinetic energy of the hammer just before impact is \( \frac{1}{2}Mv^2 = Mgh \). This energy is used to do work against the opposing force and can be related to the penetration depth \( d \). Assuming the hammer comes to rest after penetrating a distance \( d \) (and ignoring the mass \( m \) of the nail for simplicity, as it's often negligible compared to \( M \) or the problem doesn't specify how the masses interact post-collision), the kinetic energy of the hammer is converted into work done against \( F \): \[ Mgh = F \cdot d \] #### 6: Solve for Opposing Force \( F \) Rearranging to solve for \( F \): \[ F = \frac{Mgh}{d} \] ### Example Given \( M = 10 \, \text{kg} \), \( h = 2 \, \text{m} \), \( d = 0.1 \, \text{m} \): \[ F = \frac{10 \times 9.8 \times 2}{0.1} = 1960 \, \text{N} \] ### Practice Questions 1. **Easy**: A 5 kg hammer falls from a height of 3 m. If it penetrates 0.05 m into the ground, find the opposing force. - Solution: \( F = \frac{5 \times 9.8 \times 3}{0.05} = 2940 \, \text{N} \) 2. **Medium**: A 20 kg pile driver falls from a height of 5 m. It drives a pile 0.2 m into the ground. Calculate the average opposing force. - Solution: \( F = \frac{20 \times 9.8 \times 5}{0.2} = 4900 \, \text{N} \) 3. **Challenging**: If the pile driver (20 kg) falls from a height \( h \) and penetrates \( d = 0.1h \) into the ground, express the opposing force in terms of \( h \). - Solution: \( F = \frac{20 \times 9.8 \times h}{0.1h} = 1960 \, \text{N} \) ### Conclusion The opposing force against the penetration of the nail into the ground can be determined by equating the energy of the falling hammer to the work done against the opposing force during penetration. This approach provides a straightforward method to solve such problems, emphasizing the conservation of energy and the definition of work.