A3-1. Work, Energy, and Power

1. Work Done by a Constant Force

In physics, work is the effect of a force changing an object's state of motion; it is the primary measure of energy transfer. A constant force is one where both magnitude and direction remain unchanged regardless of time or position.

Calculating Constant Force Work:

Work is defined as the dot product (inner product) of the force vector and the displacement vector. It is a scalar quantity measured in Joules (J).

$$W = \vec{F} \cdot \vec{S} = F \cdot S \cdot \cos\theta$$

This formula can be interpreted in two ways:

  • Work = Magnitude of force $\times$ Component of displacement along the force direction.
  • Work = Component of force along the displacement direction $\times$ Magnitude of displacement.

Positive, Negative, and Zero Work

  • Positive Work ($W > 0$): The angle between force and displacement is $0^\circ \le \theta < 90^\circ$. The force actively contributes to the motion.
  • Negative Work ($W < 0$): The angle between force and displacement is $90^\circ < \theta \le 180^\circ$. The force opposes the motion (e.g., friction).
  • Zero Work ($W = 0$): The force is perfectly perpendicular to the displacement ($\theta = 90^\circ$). The force has absolutely no effect on the object's speed in that direction.

Example 1: Ski Jumper

Problem: A ski jumper with a mass of 50.0 kg leaves a jump ramp horizontally. The jumper lands on a slope angled at 30° to the horizontal. The distance from the launch point to the landing point is $d = 120\text{ m}$. Calculate the work done by gravity on the jumper during the flight. ($g = 9.8\text{ m/s}^2$)

30° d = 120 m mg θ=60°

Solution:

  • Analyze the vectors:
    The force of gravity is constant and acts straight downwards. The net displacement is along the 30° landing slope. The angle $\theta$ between the downward gravity vector and the displacement vector along the slope is $90^\circ - 30^\circ = 60^\circ$.
  • Calculate Work:
    $$W = F \cdot S \cdot \cos\theta = mg \cdot d \cdot \cos(60^\circ)$$
    $$W = 50.0 \times 9.8 \times 120 \times 0.5 = \mathbf{29400\text{ J}}$$
  • Insight: Because the jumper's height decreases, the angle is acute, meaning gravity does positive work, transferring potential energy into kinetic energy.

Example 2: Block on a Smooth Incline

Problem: A block of mass $m$ slides down a frictionless incline of angle $\phi$ for a distance $d$. What is the total work done by all external forces on the block?

φ N mg d

Solution:

  • 1. Work done by individual forces:
    The forces acting on the block are Gravity ($mg$) and Normal Force ($N$).
    - Gravity ($W_g$): The angle between straight downward gravity and the slope is $(90^\circ - \varphi)$. $$ W_g = mg \cdot d \cdot \cos(90^\circ - \phi) = mgd \sin\phi $$ - Normal Force ($W_N$): The normal force is perpendicular to the displacement ($\theta = 90^\circ$). $$ W_N = N \cdot d \cdot \cos(90^\circ) = 0 $$
  • 2. Total Work ($W_{total}$): $$W_{total} = W_g + W_N = \mathbf{mgd \sin\varphi}$$
  • Alternative Method: The net force acting down the slope is $F_{net} = mg \sin\varphi$. Work done by the net force is directly $W = (mg \sin\varphi) \cdot d = mgd \sin\varphi$. The sum of work done by individual forces perfectly equals the work done by the net force.

2. Work Done by a Variable Force

Unlike constant forces, variable forces change in magnitude or direction with respect to time or position. Common examples include the restoring force of a spring (which increases as it stretches) or gravitational pull over massive astronomical distances.

Graphical Method (F-x Graph):

We cannot simply use $W = F \times S$ for a variable force. Instead, we approximate the work by slicing the displacement into tiny, near-constant segments ($\Delta x$). The total work is the sum of these tiny rectangles ($\Sigma F_x \Delta x$).

In calculus terms, this perfectly matches finding the area under the Force-Position (F-x) curve.

x (Position) F (Force) Area = Total Work (W) Δx F(x)
  • Positive Work: Area enclosed above the x-axis (Force and displacement are in the positive direction).
  • Negative Work: Area enclosed below the x-axis (Force and displacement are in opposite directions).

Example 3: Work from an F-x Graph

Problem: A variable force along the x-axis acts on a particle. The area under the F-x graph from $x = 0.0\text{ m}$ to $x = 8.0\text{ m}$ is entirely above the x-axis, and from $x = 8.0\text{ m}$ to $x = 12.0\text{ m}$ it is below the x-axis. The total net work done by the force from $x = 0.0\text{ m}$ to $x = 12.0\text{ m}$ is calculated via the graph area to be 18 J.

x (m) F (N) 8.0 12.0 Positive Work Negative Work

Application Question: If the particle were to move backwards from $x = 8.0\text{ m}$ to $x = 0.0\text{ m}$, how much work is done by the force?

Solution:

  • Analyze the forward motion:
    From 0 to 8.0m, the force is positive ($F > 0$) and displacement is positive ($\Delta x > 0$), meaning the work is positive. From 8.0 to 12.0m, the force is negative ($F < 0$) but displacement is positive ($\Delta x > 0$), meaning the work is negative.
    If total work (0 to 12m) is 18 J, and the 0 to 8m section's area represents 24.0 J, then the 8 to 12m section's area is -6 J ($24 - 6 = 18$).
  • Analyze the backward motion:
    Moving from $x = 8.0\text{ m}$ to $x = 0.0\text{ m}$ means the displacement is now strictly negative ($\Delta x < 0$). However, in this region, the force is still pushing in the positive direction ($F > 0$).
    Because force and displacement are now in exact opposite directions, the work done is negative.
    Work = $\mathbf{-24.0\text{ J}}$

3. Work Done by Spring Force

Hooke's Law: $F = -kx$
The restoring force of an ideal spring is directly proportional to its deformation ($x$) and always acts in the opposite direction pointing back toward the equilibrium ($x=0$).

Work and Elastic Potential Energy:
The F-x graph for a spring is a straight line through the origin. The area of the triangle formed gives the Elastic Potential Energy:

$$E_p = \dfrac{1}{2} k x^2$$

The work done by the spring force when moving a mass from an initial position $x_i$ to a final position $x_f$ is equal to the negative change in potential energy:

$$W_s = \dfrac{1}{2} k x_i^2 - \dfrac{1}{2} k x_f^2$$

Example 4: Spring Work

Problem: A spring has a spring constant $k = 100\text{ N/m}$ and an original natural length of 0.50 m. It is placed horizontally on a smooth surface with one end fixed. An external force stretches the spring to a total length of 0.70 m. The object is then released.

  1. How much work is done by the spring force as the object returns from the stretched position back to its natural length?
  2. If the object continues moving past equilibrium to a point 0.10 m to the right of the equilibrium position, what is the total work done by the spring force from the initial release point to this new point?
Position x (m) Force |F| (N) 0 F = kx 0.10 0.20 10 20 Δ Area = 1.5 J (Work from 0.20 to 0.10) Ep = 0.5 J

Solution:

  • Part 1: Return to Equilibrium
    Initial position (deformation): $x_i = 0.70 - 0.50 = 0.20\text{ m}$
    Final position (natural length): $x_f = 0\text{ m}$
    $$W_s = \dfrac{1}{2} k x_i^2 - \dfrac{1}{2} k x_f^2 = \dfrac{1}{2}(100)(0.20)^2 - 0$$
    $$W_s = 50 \times 0.04 = \mathbf{2.0\text{ J}}$$
  • Part 2: Moving past equilibrium
    Initial position: $x_i = 0.20\text{ m}$
    Final position: $x_f = 0.10\text{ m}$
    $$W_s = \dfrac{1}{2} k x_i^2 - \dfrac{1}{2} k x_f^2 = \dfrac{1}{2}(100)(0.20)^2 - \dfrac{1}{2}(100)(0.10)^2$$
    $$W_s = 2.0\text{ J} - 0.5\text{ J} = \mathbf{1.5\text{ J}}$$
  • Insight: The formula $W_s = \frac{1}{2} k x_i^2 - \frac{1}{2} k x_f^2$ works perfectly regardless of the start and end points, automatically accounting for whether the spring is speeding the object up (positive work) or slowing it down (negative work).

4. Power

Power describes the rate at which work is done or energy is transferred. While "Work" tells us how much energy was moved, "Power" tells us how fast the transfer occurred. A high-power engine can perform a massive amount of work in a very short duration.

Average Power ($\overline{P}$): The total work done divided by the total time taken over an interval.

$$\overline{P} = \dfrac{W}{\Delta t} = \dfrac{\Delta E}{\Delta t}$$

It is measured in Watts (W). $1\text{ W} = 1\text{ J/s}$.

Instantaneous Power ($P$): The rate of doing work at a specific, precise instant in time. If a constant force moves an object at a given velocity, the instantaneous power can be expressed as:

$$P = \dfrac{\vec{F} \cdot \vec{s}}{\Delta t} = \vec{F} \cdot \vec{v} = Fv \cos\theta$$

Example 5: Average vs. Instantaneous Power of an Accelerating Car

Problem: A car with a mass of 1000 kg uniformly accelerates from rest to a speed of 12 m/s in 3.0 s. (Assume there are no resistive forces).

  1. Calculate the total work done by the net force on the car during this 3.0 s interval.
  2. Determine the average power developed by the net force over the 3.0 s.
  3. Calculate the instantaneous power exactly at $t = 2.0\text{ s}$.
t = 0 s v₀ = 0 m/s t = 2.0 s v₂ = ? F (Net) t = 3.0 s v₃ = 12 m/s Total time interval Δt = 3.0 s

Solution:

  • Part 1: Total Work Done
    Using the Work-Kinetic Energy Theorem, the total work done equals the total change in kinetic energy: $$W_{net} = \Delta E_k = \dfrac{1}{2} m v_f^2 - \dfrac{1}{2} m v_i^2$$ $$W = \dfrac{1}{2}(1000)(12)^2 - 0 = 500 \times 144 = \mathbf{72,000\text{ J}}$$
  • Part 2: Average Power over 3.0 s
    Average power is the total work divided by the total time: $$\overline{P} = \dfrac{W}{\Delta t} = \dfrac{72,000}{3.0} = \mathbf{24,000\text{ W} \text{ (or } 24\text{ kW)}}$$
  • Part 3: Instantaneous Power at $t = 2.0\text{ s}$
    To find instantaneous power ($P = F \times v$), we need the specific force and the specific velocity exactly at $t = 2.0\text{ s}$.
    a) Find the uniform acceleration:
    $$a = \dfrac{\Delta v}{\Delta t} = \dfrac{12 - 0}{3.0} = 4.0\text{ m/s}^2$$
    b) Find the net constant force:
    $$F_{net} = ma = 1000 \times 4.0 = 4000\text{ N}$$
    c) Find the velocity at $t = 2.0\text{ s}$:
    $$v = v_i + at \implies v = 0 + (4.0)(2.0) = 8.0\text{ m/s}$$
    d) Calculate Instantaneous Power:
    $$P = F \times v = 4000 \times 8.0 = \mathbf{32,000\text{ W} \text{ (or } 32\text{ kW)}}$$
    Insight: Notice how the instantaneous power at 2.0s is much higher than the average power over the whole trip. Since the car is constantly speeding up, the rate at which work is being done (power) is constantly increasing!