Problem 91
Question
A 4.00-g bullet, traveling horizontally with a velocity of magnitude \(400 \mathrm{m} / \mathrm{s},\) is fired into a wooden block with mass \(0.800 \mathrm{kg},\) initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 190 \(\mathrm{m} / \mathrm{s}\) . The block slides a distance of 45.0 \(\mathrm{cm}\) along the surface from its initial position. (a) What is the coefficient of kinetic friction between block and surface? (b) What is the decrease in kinetic energy of the bullet? (c) What is the kinetic energy of the block at the instant after the bullet passes through it?
Step-by-Step Solution
Verified Answer
(a) 0.125. (b) 247.8 J decrease. (c) 0.441 J KE of block.
1Step 1: Calculate Bullet's Initial and Final Momentum
The bullet's initial momentum is given by \( p_{i} = m_{b}v_{b} \) where \( m_{b} = 0.004\, \text{kg} \) and \( v_{b} = 400\, \text{m/s} \). Thus, \( p_{i} = 0.004 \times 400 = 1.6\, \text{kg} \cdot \text{m/s} \).\The bullet's final momentum after passing through the block is \( p_{f} = m_{b}v_{f} \) where \( v_{f} = 190\, \text{m/s} \). Thus, \( p_{f} = 0.004 \times 190 = 0.76\, \text{kg} \cdot \text{m/s} \).
2Step 2: Apply Conservation of Linear Momentum
The conservation of linear momentum during the collision implies: \( m_{b}v_{b} = m_{b}v_{f} + m_{block}v_{block} \). \Substitute the known values:\[ 0.004 \times 400 = 0.004 \times 190 + 0.800\, v_{block} \]. Solving for \( v_{block} \), we get: \\[ 1.6 = 0.76 + 0.800\, v_{block} \]\[ 0.84 = 0.800\, v_{block} \]\[ v_{block} = \frac{0.84}{0.800} = 1.05\, \text{m/s} \].
3Step 3: Calculate the Coefficient of Kinetic Friction
To find the coefficient of kinetic friction \( \mu \), use the work-energy principle. The block's kinetic energy is dissipated by friction. The work done by friction is \( W = \text{friction force} \times \text{distance} = \mu mgd \). \The kinetic energy of the block is \( KE = \frac{1}{2}m_{block}v_{block}^2 = \frac{1}{2} \times 0.800 \times (1.05)^2 \).\This gives: \[ KE = 0.441\, \text{J} \]. \Set \( KE = \mu mgd \) and solve for \( \mu \): \\( 0.441 = \mu \times (0.800 \times 9.81 \times 0.45) \) \\[ 0.441 = 3.5286 \mu \] \\[ \mu = \frac{0.441}{3.5286} = 0.125 \].
4Step 4: Calculate Decrease in Kinetic Energy of Bullet
The initial kinetic energy of the bullet is \( KE_{initial} = \frac{1}{2}m_{b}v_{b}^2 = \frac{1}{2} \times 0.004 \times 400^2 \). \This gives \( KE_{initial} = 320\, \text{J} \).\The final kinetic energy of the bullet is \( KE_{final} = \frac{1}{2} \times 0.004 \times 190^2 \). \This gives \( KE_{final} = 72.2\, \text{J} \).\The decrease in kinetic energy is \( \Delta KE = KE_{initial} - KE_{final} = 320 - 72.2 = 247.8\, \text{J} \).
5Step 5: Calculate Kinetic Energy of Block after Bullet Passes
The kinetic energy of the block after the bullet passes through is given by \( KE_{block} = \frac{1}{2}m_{block}v_{block}^2 = \frac{1}{2} \times 0.800 \times (1.05)^2 \). \The calculation yields \( KE_{block} = 0.441\, \text{J} \).
Key Concepts
Coefficient of Kinetic FrictionKinetic EnergyWork-Energy Principle
Coefficient of Kinetic Friction
The coefficient of kinetic friction is a measure of how much frictional force exists between two sliding surfaces. In this scenario, the wooden block slides on a level surface, slowing down due to frictional forces opposing its motion. This is crucial in understanding how energy is dissipated.The friction force acting on the block is proportional to both the normal force and the coefficient of kinetic friction, denoted as \( \mu \). The frictional force can be computed as:
- Friction Force \( F = \mu m_{block} g \)
Kinetic Energy
Kinetic energy is the energy an object has due to its motion, playing a crucial role in the dynamics of moving bodies. For any object in motion, its kinetic energy is calculated using the equation:
- \( KE = \frac{1}{2} m v^2 \)
- Initial bullet kinetic energy is significantly higher, reflecting its high speed.
- After impact, both bullet and block have reduced kinetic energies as energy has dissipated into the environment as thermal energy and through work done by friction on the block.
Work-Energy Principle
The work-energy principle gives a profound insight into how forces acting on an object result in changes in energy over a distance. It states that the work done by all forces acting on an object equals the change in the kinetic energy of the object. In formulaic terms, this can be expressed as:
- \( W = \Delta KE \)
- The work-energy principle helps in calculating the friction work as the product of friction force and distance.
- This principle aids in equating the initial kinetic energy of the block to the work done by friction, hence finding the friction coefficient \( \mu \).
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