Question

A 4.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.800 kg, initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 190 m/s. The block slides a distance of 72.0 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?

Short answer

(a) \( \mu_k \approx 0.078 \), (b) Decrease in KE of bullet is \( 246.1 \text{J} \), (c) KE of block is \( 0.441 \text{J} \).

Step-by-step solution

Identify Known Variables

First, list the known variables: mass of the bullet \( m_1 = 0.004 \, \text{kg} \), initial velocity of the bullet \( v_{1i} = 400 \, \text{m/s} \), final velocity of the bullet \( v_{1f} = 190 \, \text{m/s} \), mass of the block \( m_2 = 0.800 \, \text{kg} \), distance block slides \( d = 0.720 \, \text{m} \).

Use Conservation of Momentum

Apply the conservation of momentum to find the velocity of the block \( v_{2f} \) after the bullet passes through: \[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \] Since the block is initially at rest, \( v_{2i} = 0 \), so:\[ 0.004 \times 400 = 0.004 \times 190 + 0.800 \times v_{2f} \] Solve for \( v_{2f} \).

Calculate Velocity of Block

Substitute the known values into the equation from Step 2:\[ 1.6 = 0.76 + 0.800 \times v_{2f} \] \[ 0.84 = 0.800 \times v_{2f} \] \[ v_{2f} = \frac{0.84}{0.800} = 1.05 \, \text{m/s} \].

Find Kinetic Energy of Block

Immediately after the bullet passes:\[ KE_{2f} = \frac{1}{2} m_2 v_{2f}^2 \] Substitute \( m_2 = 0.800 \, \text{kg} \) and \( v_{2f} = 1.05 \, \text{m/s} \) into the equation:\[ KE_{2f} = \frac{1}{2} \times 0.800 \times (1.05)^2 \approx 0.441 \text{J} \].

Determine Kinetic Energy Decrease of Bullet

Calculate the decrease in kinetic energy of the bullet:\[ \Delta KE_1 = \frac{1}{2} m_1 (v_{1i}^2 - v_{1f}^2) \] Substitute values:\[ \Delta KE_1 = \frac{1}{2} \times 0.004 \times (400^2 - 190^2) \approx 246.1 \, \text{J} \].

Calculate Frictional Force Using Work-Energy Principle

The block's kinetic energy is dissipated due to friction. Use the work-energy principle:\[ f_k d = KE_{2f} \] Where \( f_k = \mu_k m_2 g \) and \( d = 0.72 \, \text{m} \):\[ \mu_k \times 0.800 \times 9.8 \times 0.72 = 0.441 \] Solve for \( \mu_k \).

Solve for Coefficient of Kinetic Friction

Substitute into the friction work equation:\[ \mu_k \times 5.664 = 0.441 \] \[ \mu_k = \frac{0.441}{5.664} \approx 0.078 \].

Key concepts

Kinetic Friction

When objects move across a surface, they experience a force that opposes the motion. This force is called kinetic friction. It's the reason why the wooden block in the exercise eventually comes to a stop after the bullet passes through. To calculate kinetic friction, we often use the equation:

• \( f_k = \mu_k m g \)

where \( \mu_k \) is the coefficient of kinetic friction, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)). Kinetic friction depends on both the nature of the surfaces in contact and how hard they are pressed together.
The coefficient of kinetic friction is a dimensionless number, meaning it has no units. It represents how slippery or rough a surface is. In the problem, it was calculated by setting the work done by the frictional force equal to the kinetic energy lost by the block as it slides to a stop. This allows us to solve for \( \mu_k \) given that:

• work done by friction = \( 0.441 \, \text{J} \),
• \( \mu_k \times 5.664 = 0.441 \)

Solving this gives us \( \mu_k = 0.078 \), which tells us about the friction between the block and the surface.

Kinetic Energy

Kinetic energy (KE) is the energy an object possesses due to its motion. It depends on two factors: the mass of the object and the speed at which it's moving. The equation to find kinetic energy is:

• \( KE = \frac{1}{2} m v^2 \)

where \( m \) is the mass and \( v \) is the velocity of the object.
In the problem, the bullet's initial and final kinetic energies were calculated using this formula. Initially, the bullet was traveling at 400 m/s, and after passing through the block, its speed was reduced to 190 m/s. We also calculated the kinetic energy of the block right after the bullet exits, as it begins to slide due to the velocity imparted by the bullet.
The bullet's decrease in kinetic energy can be found by comparing its initial and final energies. This change represents the energy transferred to the block and the energy lost to other factors like heat and deformation during impact. The block's kinetic energy immediately after the bullet passes through was around \( 0.441 \, \text{J} \), calculated using the velocity we found previously (\( 1.05 \, \text{m/s} \)).

Work-Energy Principle

The work-energy principle is a powerful tool in physics. It relates the work done by forces on an object to the changes in its kinetic energy. According to this principle, the change in kinetic energy of an object is equal to the work done by all forces:

• \( W = \Delta KE \)

Here, \( W \) represents the work done, and \( \Delta KE \) is the change in kinetic energy.
In the exercise, this principle helps us understand the relationship between the kinetic energy initially present in the bullet and its transformation as it passes through the block. It also helps in calculating the effect of friction as the block slides across the surface.
After the bullet passes through, the kinetic energy of the block is dissipated over the distance it slides due to friction. The work done by kinetic friction is the force times the distance (\( f_k \times d \)), which equals the kinetic energy lost by the block in motion. By setting up the equation with the known values (e.g., \( 0.441 \, \text{J} \) of kinetic energy, block mass, and distance), we can find the coefficient of kinetic friction. The work-energy principle thus directly connects forces, motion, and energy changes in a system.