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Chapter 8: Problem 91

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?

Short Answer

Expert verified
(a) 0.125. (b) 247.8 J decrease. (c) 0.441 J KE of block.

Step by step solution

01

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, \( p_{f} = 0.004 \times 190 = 0.76\, \text{kg} \cdot \text{m/s} \).
02

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} \]\[ 0.84 = 0.800\, v_{block} \]\[ v_{block} = \frac{0.84}{0.800} = 1.05\, \text{m/s} \].
03

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 \].
04

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 \( \Delta KE = KE_{initial} - KE_{final} = 320 - 72.2 = 247.8\, \text{J} \).
05

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 \( KE_{block} = 0.441\, \text{J} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \( F = \mu m_{block} g \)
where \( g \) is the acceleration due to gravity.Friction serves as a braking force on the block, converting kinetic energy into thermal energy and other forms of non-mechanical energy, eventually bringing the block to rest after it travels a certain distance. Calculating \( \mu \) involves using the work-energy principle, where the work done by friction equals the change in kinetic energy of the block.$$ KE = \mu m g d $$Concluding, the coefficient of kinetic friction here is derived by comparing the block's loss in kinetic energy to the distance slid over.
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:
  • \( KE = \frac{1}{2} m v^2 \)
where \( m \) is the mass and \( v \) is the velocity.In this exercise, both the bullet and the block experience changes in their kinetic energies due to their interaction. Initially, the bullet has a substantial amount of kinetic energy due to its high velocity, which is partly transferred to the block as it continues its journey after exiting the block. Calculating the initial and final kinetic energies of the bullet helps us determine the net energy transfer.
  • 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.
Understanding kinetic energy changes is key in analyzing collision outcomes and energy conservation across the system.
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:
  • \( W = \Delta KE \)
where \( W \) represents the work done.For the block in this scenario, the work done by friction is the primary factor reducing its kinetic energy. The block's smooth deceleration along the surface directly results from the frictional work done against its motion:
  • 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 \( \mu \).
Thus, applying the work-energy principle provides clear insights into energy transformations and conservation laws in mechanical systems.

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Most popular questions from this chapter

Force of a Baseball Swing. A baseball has mass 0.145 \(\mathrm{kg}\) . (a) If the velocity of a pitched ball has a magnitude of 45.0 \(\mathrm{m} / \mathrm{s}\) and the batted ball's velocity is 55.0 \(\mathrm{m} / \mathrm{s}\) in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. (b) If the ball remains in contact with the bat for 2.00 \(\mathrm{ms}\) , find the magnitude of the average force applied by the bat.

To warm up for a match, a tennis player hits the \(57.0-\) g ball vertically with her racket. If the ball is stationary just before it is hit and goes 5.50 \(\mathrm{m}\) high, what impulse did she impart to it?

A fireworks rocket is fired vertically upward. At its maximum height of \(80.0 \mathrm{m},\) it explodes and breaks into two pieces: one with mass 1.40 \(\mathrm{kg}\) and the other with mass 0.28 \(\mathrm{kg} .\) In the explosion, 860 \(\mathrm{J}\) of chemical energy is converted to kinetic energy of the two fragments. (a) What is the speed of each fragment just after the explosion? (b) It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.

Three odd-shaped blocks of chocolate have the following masses and center-of- mass coordinates: \((1) 0.300 \mathrm{kg},(0.200 \mathrm{m},\) \(0.300 \mathrm{m} ) ;\) (2) \(0.400 \mathrm{kg},(0.100 \mathrm{m},-0.400 \mathrm{m}) ;\) (3) 0.200 \(\mathrm{kg}\) \((-0.300 \mathrm{m}, 0.600 \mathrm{m}) .\) Find the coordinates of the center of mass of the system of three chocolate blocks.

CALC Starting at \(t=0,\) a horizontal net force \(\vec{F}=\) \((0.280 \mathrm{N} / \mathrm{s}) \hat{t} \hat{\imath}+\left(-0.450 \mathrm{N} / \mathrm{s}^{2}\right) t^{2} \hat{\mathrm{J}}\) is applied to a box that has an initial momentum \(\vec{p}=(-3.00 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}) \hat{\imath}+(4.00 \mathrm{kg} \cdot \) \(\mathrm{m} / \mathrm{s} ) \hat{\boldsymbol{J}}\) . What is the momentum of the box at \(t=2.00 \mathrm{s} ?\)
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