To solve this problem, we will break it down step by step using the principles of energy conservation and kinematics.

Key Information from the Problem:

• Mass of the block $m = 414 \, \text{g} = 0.414 \, \text{kg}$
• Friction coefficient $\mu = 0.2$
• The rough patch is $1 \, \text{m}$ long
• Acceleration due to gravity $g = 9.81 \, \text{m/s}^2$
• Horizontal distance (flight) $4.6 \, \text{m}$
• The vertical height of the fall after leaving the track $2.2 \, \text{m}$

We need to find the release height $h$ above the ground.

Step 1: Conservation of Energy

At the starting point (height $h$), the block has potential energy and no kinetic energy because it is released from rest.

Total mechanical energy at the top: $E_{\text{initial}} = mgh$ where $h$ is the unknown height.

At the point right before the block reaches the rough surface, all the initial potential energy has converted into kinetic energy (since the track is frictionless before that point): $E_{\text{initial}} = \frac{1}{2}mv^2$

Step 2: Work Done by Friction

When the block moves over the rough surface of length $1 \, \text{m}$, friction does negative work on the block, reducing its kinetic energy.

The work done by friction $W_{\text{friction}}$ is: $W_{\text{friction}} = \mu mg d$ where:

• $\mu = 0.2$ (coefficient of friction)
• $d = 1 \, \text{m}$ (distance over the rough patch)

The reduction in kinetic energy due to friction: $\Delta E = W_{\text{friction}} = \mu mg d = 0.2 \times 0.414 \times 9.81 \times 1 = 0.812 \, \text{J}$

Step 3: Kinematics of the Flight

The block leaves the track with some horizontal velocity $v$ and falls a vertical height of $2.2 \, \text{m}$. Using the equation of motion for free fall, we can find the time it takes for the block to hit the ground.

The vertical displacement equation is: $y = \frac{1}{2}gt^2$ Solving for $t$ (time of flight): $2.2 = \frac{1}{2} \times 9.81 \times t^2 \quad \Rightarrow \quad t = \sqrt{\frac{2 \times 2.2}{9.81}} \approx 0.67 \, \text{seconds}$

Step 4: Horizontal Velocity

The block travels $4.6 \, \text{m}$ horizontally in the same time $t$. Thus, the horizontal velocity $v$ is: $v = \frac{4.6}{t} = \frac{4.6}{0.67} \approx 6.87 \, \text{m/s}$

Step 5: Kinetic Energy at the End of the Rough Patch

At the point where the block leaves the rough patch, it still has kinetic energy: $\frac{1}{2}mv^2 = \frac{1}{2} \times 0.414 \times (6.87)^2 \approx 9.77 \, \text{J}$

Step 6: Energy Conservation to Find $h$

Before the block reaches the rough patch, its total mechanical energy was the initial potential energy minus the energy lost to friction: $mgh - \Delta E = 9.77 \, \text{J}$ Substituting the known values: $0.414 \times 9.81 \times h - 0.812 = 9.77$ Solving for $h$: $4.06h = 10.582 \quad \Rightarrow \quad h \approx 2.61 \, \text{m}$

Thus, the release height $h$ is approximately 2.61 meters.

Would you like a detailed breakdown of any specific steps or further explanation? Here are some related questions:

1. What is the total mechanical energy of the system at different points?
2. How does friction affect the kinetic energy of the block?
3. How is the time of flight derived from the vertical motion equation?
4. How would the result change if the rough patch were longer or had a different friction coefficient?
5. Can you apply the same method to a different track shape or release angle?

Tip: Always consider energy conservation when solving motion problems involving gravity and friction.