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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

\(\mathrm{a}\) (a) A 75-kg man steps out a window and falls (from rest) \(1.0 \mathrm{~m}\) to a sidewalk. What is his speed just before his feet strike the pavement? (b) If the man falls with his knees and ankles locked, the only cushion for his fall is an approximately \(0.50-\mathrm{cm}\) give in the pads of his feet. Calculate the average force exerted on him by the ground in this situation. This average force is sufficient to cause damage to cartilage in the joints or to break bones.

A horizontal spring attached to a wall has a force constant of \(850 \mathrm{~N} / \mathrm{m} . \mathrm{A}\) block of mass \(1.00 \mathrm{~kg}\) is attached to the spring and oscillates freely on a horizontal, frictionless surface as in Active Figure \(5.20 .\) The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? (b) What are the two points of interest? (c) Find the energy stored in the spring when the mass is stretched \(6.00 \mathrm{~cm}\) from equilibrium and again when the mass passes through cquilibrium after being released from rest. (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. (e) What is the speed at the halfway point? Why isn't it half the speed at equilibrium?

In 1990 Walter Arfeuille of Belgium lifted a \(281.5-\mathrm{kg}\) object through a distance of \(17.1 \mathrm{~cm}\) using only his teeth. (a) How much work did Arfeuille do on the object? (b) What magnitude force did he exert on the object during the lift, assuming the force was constant?

A truck travels uphill with constant velocity on a highway with a \(7.0^{\circ}\) slope. A 50 -kg package sits on the floor of the back of the truck and does not slide, due to a static frictional force. During an interval in which the truck travels \(340 \mathrm{~m}\), what is the net work done on the package? What is the work done on the package by the force of gravity, the normal force, and the friction force?

A \(60.0\) -kg athlete leaps straight up into the air from a trampoline with an initial speed of \(9.0 \mathrm{~m} / \mathrm{s}\). The goal of this problem is to find the maximum height she attains and her speed at half maximum height. (a) What are the interacting objects and how do they interact? (b) Select the height at which the athlete's speed is \(9.0 \mathrm{~m} / \mathrm{s}\) as \(y=0 .\) What is her kinetic energy at this point? What is the gravitational potential energy associated with the athlete? (c) What is her kinetic energy at maximum height? What is the gravitational potential energy associated with the athlete? (d) Write a general equation for energy conservation in this case and solve for the maximum height. Substitute and obtain a numerical answer. (c) Write the general equation for energy conservation and solve for the velocity at half the maximum height. Substitute and obtain a numerical answer.
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