Physics for Scientists and Engineers: Work and Energy Study Notes

Work Done by a Constant Force

• Definition of Work: The work done by a constant force is defined as the product of the distance moved and the component of the force in the direction of that displacement. The formula is expressed as:     $W = F d \cos(\theta)$     where $F$ is the magnitude of the constant force, $d$ is the displacement (distance moved), and $\theta$ is the angle between the force vector and the displacement vector.

• Units of Work: In the International System of Units (SI), work is measured in Joules ($J$).     $1\,J = 1\,N \cdot m$

• Conceptual Application (Direction of Force): If a person carries a bag of groceries horizontally at a constant height, they are doing no work on the bag. This is because the force exerted by the person (upward) is perpendicular (at a $90^\circ$ angle) to the direction of motion (horizontal). Since $\cos(90^\circ) = 0$, the work done is zero.

• Example 7-1: Work Done on a Crate:

  • Scenario: A person pulls a $50\,kg$ crate for a distance of $40\,m$ along a horizontal, smooth floor. The constant force applied is $F_P = 100\,N$ at an angle of $37^\circ$ relative to the floor.

  • Requirements: Determine (a) the work done by each force acting on the crate, and (b) the net work done on the crate.

  • Forces involved: Tension/Pulling force ($F_P$), Gravity ($mg$), Normal force ($F_N$).

• Methodology for Solving Work Problems:

  1. Draw a free-body diagram to identify all forces.

  2. Choose an appropriate coordinate system.

  3. Apply Newton’s laws to find any unknown force magnitudes if necessary.

  4. Calculate the work done by each specific force individually.

  5. Calculate net work ($W_{net}$) using one of two methods:

     • Find the net force ($F_{net}$) and calculate its work: $W_{net} = F_{net} d \cos(\theta)$.

     • Sum the work done by every individual force: $W_{net} = W_1 + W_2 + ... + W_n$.

• Example 7-2: Work on a Backpack:

  • Scenario: A hiker carries a $15.0\,kg$ backpack up a hill with a vertical height $h = 10.0\,m$. Assume smooth motion at a constant velocity (acceleration is zero).

  • Requirements: Calculate (a) the work done by the hiker on the backpack, (b) the work done by gravity on the backpack, and (c) the net work done on the backpack.

• Conceptual Example 7-3: Earth and Moon:

  • Scenario: The Moon revolves around the Earth in a nearly circular orbit at an approximately constant tangential speed, held by Earth's gravitational pull.

  • Question: Does gravity do positive, negative, or no work on the Moon?

  • Logic: Because the gravitational force is centripetal (directed toward the center of the Earth) and the displacement is tangential, the angle between the force and displacement is always $90^\circ$. Therefore, gravity does no work on the Moon.

Scalar Product of Two Vectors

• Definition of the Scalar Product: Also known as the dot product, the scalar product of two vectors $\mathbf{A}$ and $\mathbf{B}$ is defined as:     $\mathbf{A} \cdot \mathbf{B} = AB \cos(\theta)$     where $A$ and $B$ are the magnitudes of the vectors and $\theta$ is the angle between them.

• Work in Scalar Product Form: Work can be compactly written as the dot product of the force vector and the displacement vector:     $W = \mathbf{F} \cdot \mathbf{d}$

• Example 7-4: Using the Dot Product:

  • Scenario: A wagon is dragged $100\,m$ along the ground by a force with a magnitude of $F_P = 20\,N$ at an angle of $30^\circ$ to the ground.

  • Calculation: Use the dot product formula to determine the work done by the force: $W = (20\,N)(100\,m) \cos(30^\circ)$.

Work Done by a Varying Force

• General Concept: If a force varies as a particle moves, the standard product $F \cdot d$ cannot be used because $F$ is not constant. Instead, the displacement is divided into infinitesimally small pieces ($dx$).

• Approximation: The total work is approximated by summing the work done over many small intervals:     $W \approx \sum F_i \Delta x_i$

• Integral Form: In the limit that these intervals become infinitesimally narrow, the work is defined as the integral of the force over the distance:     $W = \int_{x_1}^{x_2} \mathbf{F}(x) \cdot d\mathbf{x}$

• Graphical Interpretation: On a plot of Force ($F$) versus position ($x$), the work done is equivalent to the area under the curve between the initial and final positions.

• Work Done by a Spring Force:

  • Force Equation: According to Hooke's Law, the force exerted by a spring is $F_s = -kx$, where $k$ is the spring constant and $x$ is the displacement from the equilibrium position.

  • The work done by a person to stretch or compress a spring (applying a force $F_{app} = kx$) is:         $W = \int_0^x kx \, dx = \frac{1}{2} kx^2$

• Example 7-5: Work Done on a Spring:

  • Part (a): A person stretches a spring $3.0\,cm$ ($0.030\,m$), requiring a maximum force of $75\,N$. Calculate the work done.

  • Part (b): Calculate the work done if the person compresses the spring by the same distance ($3.0\,cm$).

• Example 7-6: Force as a Function of Position:

  • Scenario: A motor manipulates a robot arm controlling a video camera. The motor exerts a force given as a function of position $x$.

  • Constants provided: $F_0 = 2.0\,N$, $x_0 = 0.0070\,m$.

  • Calculation: Determine the work done by the motor as the arm moves from $x_1 = 0.010\,m$ to $x_2 = 0.050\,m$.

Kinetic Energy and the Work-Energy Principle

• Definition of Energy: Traditionally, energy was defined as the "ability to do work." In mechanical systems, this definition remains relevant.

• Defining Kinetic Energy ($K$): By relating acceleration to velocity and distance through Newton's laws and kinematics, we derive kinetic energy as the energy of motion:     $K = \frac{1}{2}mv^2$

• The Work-Energy Principle: The net work done on an object is equal to the change in the object's kinetic energy:     $W_{net} = \Delta K = K_2 - K_1 = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

  • If W_{net} > 0, the kinetic energy increases.

  • If W_{net} < 0, the kinetic energy decreases.

• Units: Since work and kinetic energy are equated, kinetic energy is also measured in Joules ($J$).

• Example 7-7: Kinetic Energy and Work on a Baseball:

  • Scenario: A $145\,g$ ($0.145\,kg$) baseball is thrown at a speed of $25\,m/s$.

  • Requirements: (a) Calculate its kinetic energy. (b) Determine the net work done to accelerate the ball from rest to this speed.

• Example 7-8: Work on a Car (Increasing KE):

  • Scenario: A $1000\,kg$ car is accelerated from $20\,m/s$ to $30\,m/s$.

  • Requirement: Calculate the net work required for this acceleration.

• Example 7-9: Work to Stop a Car:

  • Scenario: A car traveling at $60\,km/h$ can brake to a stop in a distance $d = 20\,m$.

  • Question: If the car travels twice as fast ($120\,km/h$), what is its stopping distance, assuming the maximum braking force remains independent of speed?

  • Note: Because stopping distance $d$ is proportional to the square of the velocity ($v^2$) in the work-energy theorem, doubling the speed results in four times the stopping distance ($80\,m$).

• Example 7-10: A Compressed Spring and Block:

  • Scenario: A horizontal spring has a constant $k = 360\,N/m$.

  • Part (a): Work required to compress it $11.0\,cm$ ($0.110\,m$) from equilibrium ($x=0$).

  • Part (b): A $1.85\,kg$ block is placed against the compressed spring. Determine the speed of the block when it separates from the spring at $x = 0$ (ignore friction).

  • Part (c): Repeat part (b), but assume a constant drag force (friction) of $F_D = 7.0\,N$ acts on the block while it is in motion.

Chapter Summary

• Work done by a constant force: $W = F d \cos(\theta)$.

• Work done by a variable force: $W = \int_{x_1}^{x_2} F(x) \, dx$.

• Kinetic Energy: $K = \frac{1}{2}mv^2$.

• Work-Energy Principle: $W_{net} = \Delta K$. The net work done on an object equals the change in its kinetic energy.