Let's solve the problem step by step.
Part 1: Determine the elastic energy U stored in the compressed spring.
The elastic potential energy stored in a compressed spring can be calculated using Hooke’s law: \[ U = \frac{1}{2} k x^2 \] where:
- \( U \) is the elastic potential energy,
- \( k \) is the spring constant (632 N/m),
- \( x \) is the compression distance (0.2 m).
Substituting the values: \[ U = \frac{1}{2} (632 , \text{N/m}) (0.2 , \text{m})^2 \] \[ U = \frac{1}{2} (632) (0.04) = \frac{1}{2} (25.28) = 12.64 , \text{J} \]
Answer for Part 1:
\[ U = 12.64 , \text{J} \]
Part 2: Find the speed of the 8 kg block just before it collides with the 3 kg block.
When the spring is released, the potential energy stored in the spring is converted into kinetic energy and work done against friction.
First, we calculate the work done against friction before the block moves: The force of friction \( f_f = \mu m g \), where:
- \( \mu = 0.4 \) (coefficient of friction)
- \( m = 8 , \text{kg} \) (mass of the 8 kg block)
- \( g = 9.8 , \text{m/s}^2 \)
Calculating the force of friction: \[ f_f = \mu mg = 0.4 \times 8 , \text{kg} \times 9.8 , \text{m/s}^2 = 31.36 , \text{N} \]
Now, let's find the acceleration \( a \) of the block due to friction: \[ f_f = ma \quad \Rightarrow \quad 31.36 , \text{N} = 8 , \text{kg} \times a \] \[ a = \frac{31.36 , \text{N}}{8 , \text{kg}} = 3.92 , \text{m/s}^2 \]
Now find the net energy after accounting for work done against friction. The initial energy is entirely from the spring: Initial energy from spring: \[ U = 12.64 , \text{J} \]
Work done against friction while moving \( d \) meters (from compression \( 0.2 , \text{m} \)): \[ W_f = f_f \cdot d = 31.36 \cdot 0.2 = 6.272 , \text{J} \]
Thus, the effective kinetic energy \( K \) just before collision is: \[ K = U - W_f = 12.64 , \text{J} - 6.272 , \text{J} = 6.368 , \text{J} \]
The kinetic energy is also given by: \[ K = \frac{1}{2} mv^2 \] where \( m = 8 , \text{kg} \) and we solve for \( v \): \[ 6.368 = \frac{1}{2} \cdot 8 \cdot v^2 \] \[ 6.368 = 4 v^2 \quad \Rightarrow \quad v^2 = \frac{6.368}{4} = 1.592 \quad \Rightarrow \quad v = \sqrt{1.592} \approx 1.26 , \text{m/s} \]
Answer for Part 2:
\[ v \approx 1.26 , \text{m/s} \]
Part 3: Find the final speed of both blocks (stuck together) just after they collide.
Using the conservation of momentum: Let \( v_i = 1.26 , \text{m/s} \) (speed of the 8 kg block before the collision) and assume the 3 kg block is at rest initially (speed = 0).
\[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) V_f \] where:
- \( m_1 = 8 , \text{kg} \)
- \( m_2 = 3 , \text{kg} \)
- \( v_2 = 0 \)
Substituting the values: \[ (8 \cdot 1.26) + (3 \cdot 0) = (8 + 3)V_f \] \[ 10.08 = 11 V_f \quad \Rightarrow \quad V_f = \frac{10.08}{11} \approx 0.916 , \text{m/s} \]
Answer for Part 3:
\[ V_f \approx 0.916 , \text{m/s} \]
Part 4: Find the horizontal distance the blocks move before coming to rest.
We need to consider the friction acting on the combined mass of the two blocks. The total mass: \[ m = m_1 + m_2 = 8 , \text{kg} + 3 , \text{kg} = 11 , \text{kg} \]
The force of friction \( f_f = \mu m g \): \[ f_f = 0.4 \times 11 \times 9.8 = 43.12 , \text{N} \]
The acceleration \( a \) (deceleration here) due to friction: \[ a = \frac{f_f}{m} = \frac{43.12}{11} \approx 3.92 , \text{m/s}^2 \]
Now, using the equation of motion to find distance \( d \) to come to rest from velocity \( V_f \): \[ V_f^2 = u^2 + 2ad \] where \( u = 0.916 , \text{m/s} \) and \( a = -3.92 , \text{m/s}^2 \) (since it's deceleration).
Rearranging gives: \[ 0 = (0.916)^2 + 2(-3.92)d \] \[ 0 = 0.839056 - 7.84d \] \[ 7.84d = 0.839056 \quad \Rightarrow \quad d = \frac{0.839056}{7.84} \approx 0.107 , \text{m} \]
Answer for Part 4:
\[ d \approx 0.107 , \text{m} \]
In summary:
- Total elastic energy stored: 12.64 J
- Speed of the 8 kg block right before the collision: 1.26 m/s
- Final speed after collision (both blocks together): 0.916 m/s
- Horizontal distance before coming to rest: 0.107 m
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