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Chapter 5: Problem 15

A \(7.80\) -g bullet moving at \(575 \mathrm{~m} / \mathrm{s}\) penetrates a tree trunk to a depth of \(5.50 \mathrm{~cm}\). (a) Use work and energy considerations to find the average frictional force that stops the bullet. (b) Assuming the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment it stops moving.

Short Answer

Expert verified
The average frictional force that stops the bullet is approximately 23539.09 N and the time it takes to stop the bullet after it has entered the tree is approximately \(1.9*10^-4\) seconds.

Step by step solution

01

Define Given Variables

Identify and define the given variables. The mass of the bullet \(m=7.80\) g which is \(0.00780\) kg when converted to standard SI units, initial speed \(v_i=575\) m/s, final speed \(v_f=0\) m/s (because the bullet stops), and depth penetrated into the tree \(d=5.5\) cm which is \(0.055\) m in meters.
02

Average Frictional Force (Part a)

According to the work-energy theorem, the work done \(W\) on the bullet by the frictional force is equal to the change in the bullet's kinetic energy. The initial kinetic energy is \(\frac{1}{2} m v_i^2\) and the final kinetic energy is \(\frac{1}{2} m v_f^2\), therefore, \(W = 螖KE = \frac{1}{2} m v_i^2 - \frac{1}{2} m v_f^2\). We find \(W = 1294.65\) J. The work done by the frictional force is also equal to the force times the distance over which the force is applied: \(W = F_f * d\). We can solve for the frictional force, giving us \(F_f = \frac{W}{d} = 23539.09\) N.
03

Time of Motion (Part b)

Assuming the frictional force is constant, the bullet experiences a constant acceleration (actually a deceleration, because it is slowing down). Therefore, we can use the kinematic equation \(v_f = v_i + a*t\) to solve for \(t\). We first need to calculate the deceleration using the formula \(a = \frac{F}{m} = -3017720.51 \mathrm{~m/s^2}\) (negative because it is slowing down). Then we substitute our known quantities into the kinematic equation and solve for \(t\), yielding \(t = \frac{v_f - v_i}{a} = 1.9*10^-4\) s.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a type of energy that an object possesses due to its motion. It is one of the key concepts in physics, especially when dealing with moving objects. The formula to calculate kinetic energy, denoted as \( KE \), is:
  • \( KE = \frac{1}{2} m v^2 \)
Here, \( m \) is the mass of the object and \( v \) is its velocity.
The kinetic energy is measured in joules (J) when the mass is in kilograms (kg) and velocity is in meters per second (m/s).
In our problem, the bullet had a significant initial velocity, leading to high kinetic energy at the start. This energy is key to understanding how the bullet manages to penetrate the tree.
Frictional Force
Frictional force is a force that opposes the relative motion or tendency of such motion of two surfaces in contact. It plays a crucial role in stopping the bullet.
  • Work is done by the frictional force as the bullet penetrates the tree.
  • The work-energy theorem relates the work done by friction to the change in kinetic energy.
In this context, the frictional force can be computed by setting it equal to the work done to stop the bullet. The calculation is made using the formula:
  • \( W = F_f \cdot d \)
By rearranging this, we find that the frictional force \( F_f \) is calculated as:
  • \( F_f = \frac{W}{d} \)
This force is remarkably high, showing how much energy is transferred from the bullet to the tree to stop it.
Constant Acceleration
Constant acceleration occurs when an object鈥檚 velocity changes at a consistent rate over time. In the scenario of the bullet, it undergoes what we call deceleration due to a constant negative force (friction).
  • The deceleration can be derived from the frictional force acting on the bullet.
  • Using \( a = \frac{F}{m} \), we observe the high deceleration as the bullet slows down.
This deceleration is important as it helps calculate how long the bullet takes to stop.
Kinematic Equations
Kinematic equations are used to describe motion under constant acceleration. They allow us to relate different aspects of motion like velocity, acceleration, and time.
To solve for time, we employ the kinematic equation:
  • \( v_f = v_i + a \cdot t \)
Where:
  • \( v_f \) is the final velocity (0 m/s in this case).
  • \( v_i \) is the initial velocity.
  • \( a \) is the acceleration.
  • \( t \) is time.
Rearranging to solve for time, \( t \), becomes:
  • \( t = \frac{v_f - v_i}{a} \)
This approach helps determine the very short time interval in which the bullet comes to a stop.

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Most popular questions from this chapter

A block of mass \(2.50 \mathrm{~kg}\) is pushed \(2.20 \mathrm{~m}\) along a frictionless horizontal table by a constant \(16.0-\mathrm{N}\) force directed \(25.0^{\circ}\) below the horizontal. Determine the work done by (a) the applied force, (b) the normal force exerted by the table, (c) the force of gravity, and (d) the net force on the block.

A 2 \(300-\mathrm{kg}\) pile driver is used to drive a steel beam into the ground. The pile driver falls \(7.50 \mathrm{~m}\) before coming into contact with the top of the beam, and it drives the beam \(18.0 \mathrm{~cm}\) farther into the ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.

Find the height from which you would have to drop a ball so that it would have a speed of \(9.0 \mathrm{~m} / \mathrm{s}\) just before it hits the ground.

Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is 1 kilocalorie, which we define in Chapter 11 as \(1 \mathrm{kcal}=4186 \mathrm{~J}\). Metabolizing 1 gram of fat can release \(9.00 \mathrm{kcal}\), A student decides to try to lose weight by exercising. She plans to run up and down the stairs in a football stadium as fast as she can and as many times as necessary. Is this in itself a practical way to lose weight? To evaluate the program, suppose she runs up a flight of 80 steps, each \(0.150 \mathrm{~m}\) high, in \(65.0 \mathrm{~s}\). For simplicity, ignore the energy she uses in coming down (which is small). Assume that a typical efficiency for human muscles is \(20.0 \%\). This means that when your body converts \(100 \mathrm{~J}\) from metabolizing fat, \(20 \mathrm{~J}\) goes into doing mechanical work (here, climbing stairs). The remainder goes into internal energy. Assume the student's mass is \(50.0 \mathrm{~kg}\). (a) How many times must she run Lhe flight of stairs to lose 1 pound of fat? (b) What is her average power output, in watts and in horsepower, as she is running up the stairs?

(a) A block with a mass \(m\) is pulled along a horizontal surface for a distance \(x\) by a constant force \(\overrightarrow{\mathbf{F}}\) at an angle \(\theta\) with respect to the horizontal. The coefficient of kinetic friction between block and table is \(\mu_{k} .\) Is the force exerted by friction equal to \(\mu_{h} m g_{2}^{2}\) If not, what is the force exerted by friction? (b) How much work is done by the friction force and by \(\vec{F}\) ? (Don't forget the signs.) (c) Identify all the forces that do no work on the block. (d) Let. \(m=2.00 \mathrm{~kg}, x=4.00 \mathrm{~m}, \theta=37.0^{\circ}, F=15.0 \mathrm{~N}\), and \(\mu_{k}=\) \(0.400\), and find the answers to parts (a) and (b).
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