Work and Kinetic Theorem: Principles, Problems, and Solutions

Fundamental Laws and Concepts of Work, Energy, and Power

• Definition of Work Done by a Constant Force:

  • Work is defined as the product of the magnitude of displacement and the component of force parallel to that displacement.
  • When a constant force vector $\vec{F}$ acts on a body and displaces it through a displacement vector $\vec{s}$ where $\theta$ represents the angle between the force and displacement vectors, the work done is given by:         $W = F s \cos(\theta)$
  • $F$ is the magnitude of the constant force.
  • $s$ is the magnitude of the displacement.
  • $\theta$ is the angle between the direction of the force and the direction of displacement.
  • Work is a scalar quantity, and its SI unit is the joule ($\text{J}$), where $1\,\text{J} = 1\,\text{N}\cdot\text{m} = 1\,\text{kg}\,\text{m}^2/\text{s}^2$.

• Work Done by a Varying Force:

  • When the applied force varies continuously with position along the direction of motion, work is determined through integration over the displacement interval from $x_1$ to $x_2$:         $W = \int_{x_1}^{x_2} F(x)\,dx$
  • $F(x)$ represents the position-dependent force function.
  • Geometrically, the work done corresponds exactly to the definite integral or the net area bounded under the curve on an $F-x$ graph between $x_1$ and $x_2$.
  • Areas located above the displacement axis represent positive work, while areas below the axis represent negative work.
  • This formulation reduces to the constant force relation $W = F s \cos(\theta)$ when $F(x)$ is uniform over the path.

• Work Done by an Ideal Spring:

  • An ideal spring exerts a restoring force governed by Hooke's Law:         $F_s = -k x$
  • $k$ denotes the spring force constant (stiffness) expressed in newtons per meter ($\text{N/m}$).
  • $x$ represents the displacement or elongation/compression relative to the unstretched equilibrium position ($x = 0$).
  • The work done by the spring force on an attached object transitioning from an initial coordinate $x_i$ to a final coordinate $x_f$ is calculated by:         $W_s = \int_{x_i}^{x_f} (-k x)\,dx = \left[ -\frac{1}{2} k x^2 \right]_{x_i}^{x_f} = \frac{1}{2} k x_i^2 - \frac{1}{2} k x_f^2$
  • Work performed against or by the spring force relates directly to the change in elastic potential energy ($U_s$), defined as:         $U = \frac{1}{2} k x^2$
  • $W_s = -\Delta U = U_i - U_f$

• The Work-Kinetic Energy Theorem:

  • The net work ($W_{\text{net}}$) performed on a particle by the vector sum of all forces equals the change in the particle's kinetic energy ($\Delta K$):         $W_{\text{net}} = \Delta K = K_f - K_i = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$
  • $m$ is the mass of the particle.
  • $v_i$ is the initial speed, and $v_f$ is the final speed.
  • This theorem applies universally to both constant and varying forces along linear or curved trajectories, bridging kinematic quantities directly to dynamical forces.

• Power and Energy Delivery:

  • Power ($P$) is defined as the time rate at which work is performed or mechanical energy is transferred:         $P = \frac{W}{t}$
  • The SI unit of power is the watt ($\text{W}$), where $1\,\text{W} = 1\,\text{J/s} = 1\,\text{kg}\,\text{m}^2/\text{s}^3$.
  • For a constant force acting in the direction of continuous motion, power is expressed as the product of force and velocity:         $P = F v$

Work Under Gravity and Constant Applied Forces

• Free-Falling Hammer Impact:

  • Problem Statement: A hammer head of mass $m = 1.2\,\text{kg}$ falls onto a nail from a height of $d = 0.50\,\text{m}$. Determine the maximum work it could do on the nail.
  • Analysis: The maximum possible work corresponds to the work performed by gravity over the fall distance, assuming all acquired energy transfers to the nail.
  • Both the gravitational force ($F_G = m g$) and the displacement ($d$) act downward in the same direction, yielding an angle $\theta = 0^\circ$.
  • Using the work formula for a constant force:         $W_G = m g d \cos(\theta)$$W_G = (1.2\,\text{kg})(9.80\,\text{m/s}^2)(0.50\,\text{m})\cos(0^\circ) = 5.88\,\text{J} \approx 5.9\,\text{J}$

• Vertical Lifting at Constant Velocity:

  • Problem Statement: Determine the work done by a person lifting an object of mass $m = 2.0\,\text{kg}$ from the bottom of a well at a constant speed of $v = 2.0\,\text{m/s}$ for a duration of $t = 5.0\,\text{s}$.
  • Displacement Calculation:         $d = v t = (2.0\,\text{m/s})(5.0\,\text{s}) = 10.0\,\text{m}$
  • Applied Force Calculation: Since speed is constant ($a = 0$), the net vertical force is zero. The lifting force ($F_{\text{lift}}$) balances gravity:         $F_{\text{lift}} = m g = (2.0\,\text{kg})(9.8\,\text{m/s}^2) = 19.6\,\text{N}$
  • Work Evaluation:         $W = F_{\text{lift}} d \cos(0^\circ) = (19.6\,\text{N})(10.0\,\text{m})(1) = 196\,\text{J} = 0.196\,\text{kJ} \approx 0.20\,\text{kJ}$

Equilibrium and Work on an Inclined Plane

• System Setup for a Restrained Piano on an Incline:
  • A piano with mass $m = 380\,\text{kg}$ slides down a frictionless incline inclined at $\theta = 25^\circ$ over a displacement of $d = 2.9\,\text{m}$.
  • A worker pushes back parallel to the incline with a force $\vec{F}_P$ to keep the piano sliding at a constant velocity ($a = 0$).

• Force Equilibrium Analysis (Newton's Second Law):

  • Establish coordinate axes with the positive $x$-axis oriented down the incline parallel to motion, and the positive $y$-axis perpendicular to the inclined plane.
  • Sum of forces in the $y$-direction:         $\sum F_y = F_N - m g \cos(\theta) = 0$$F_N = m g \cos(\theta)$
  • Sum of forces in the $x$-direction:         $\sum F_x = m g \sin(\theta) - F_P = 0$$F_P = m g \sin(\theta)$
  • Force Exerted by the Man:         $F_P = (380\,\text{kg})(9.80\,\text{m/s}^2)\sin(25^\circ) = 1573.9\,\text{N} \approx 1600\,\text{N}$

• Work Components and Net Work Evaluation:

  • Work Performed by the Man ($W_P$):
    • The pushing force acts up the plane while displacement is down the plane ($\theta = 180^\circ$).         $W_P = F_P d \cos(180^\circ) = (1573.9\,\text{N})(2.9\,\text{m})(-1) = -4564.3\,\text{J} \approx -4600\,\text{J}$
  • Work Performed by Gravity ($W_G$):
    • The angle between downward gravity ($m\vec{g}$) and the displacement along the $25^\circ$ slope is $90^\circ - 25^\circ = 65^\circ$.         $W_G = m g d \cos(65^\circ)$$W_G = (380\,\text{kg})(9.80\,\text{m/s}^2)(2.9\,\text{m})\cos(65^\circ) = 4564.3\,\text{J} \approx 4600\,\text{J}$
    • Alternatively, using gravitational potential energy change: $W_G = m g h = m g (d \sin(25^\circ)) = (380)(9.80)(2.9\sin(25^\circ)) = 4564.3\,\text{J}$.
  • Net Work Done on the Piano ($W_{\text{net}}$):
    • Because the piano slides with constant velocity, net acceleration is zero, requiring net force to be zero ($\vec{F}_{\text{net}} = 0$), which dictates:             $W_{\text{net}} = 0\,\text{J}$
    • Summing individual work contributions verifies this result:             $W_{\text{net}} = W_P + W_G = -4600\,\text{J} + 4600\,\text{J} = 0\,\text{J}$

Graphical Evaluation of Work Done by Variable Forces

• Trapezoidal Integration of Piecewise Linear Functions:
  • Problem Statement: A force $F_x$ acting along the $x$-axis varies with position as plotted above. Determine the work done from (a) $x = 0.0\,\text{m}$ to $x = 10.0\,\text{m}$ and (b) $x = 0.0\,\text{m}$ to $x = 15.0\,\text{m}$.
  • Region 1: Work from $x = 0.0\,\text{m}$ to $x = 10.0\,\text{m}$:
    • The geometric shape bounded between $x = 0$ and $x = 10.0\,\text{m}$ is a trapezoid above the horizontal axis with a maximum force height of $h = 400\,\text{N}$.
    • Bottom base ($b_1$): $10.0 - 0.0 = 10.0\,\text{m}$.
    • Top flat base ($b_2$): extending from $x = 3.0\,\text{m}$ to $x = 7.0\,\text{m}$, giving $7.0 - 3.0 = 4.0\,\text{m}$.
    • Using the trapezoidal area formula:             $W_{0 \rightarrow 10} = \frac{1}{2} (b_1 + b_2) h = \frac{1}{2} (10.0\,\text{m} + 4.0\,\text{m})(400\,\text{N}) = 2800\,\text{J}$
  • Region 2: Work from $x = 10.0\,\text{m}$ to $x = 15.0\,\text{m}$:
    • Between $x = 10.0\,\text{m}$ and $x = 15.0\,\text{m}$, the force is negative, forming a trapezoid below the horizontal axis with an altitude of $h = -200\,\text{N}$.
    • Long base ($b_1'$): extending from $x = 10.0\,\text{m}$ to $x = 15.0\,\text{m}$, giving $15.0 - 10.0 = 5.0\,\text{m}$.
    • Short flat base ($b_2'$): extending from $x = 12.0\,\text{m}$ to $x = 14.0\,\text{m}$, giving $14.0 - 12.0 = 2.0\,\text{m}$.
    • Evaluating the negative work area:             $W_{10 \rightarrow 15} = \frac{1}{2} (b_1' + b_2') h = \frac{1}{2} (5.0\,\text{m} + 2.0\,\text{m})(-200\,\text{N}) = -700\,\text{J}$
  • Total Work from $x = 0.0\,\text{m}$ to $x = 15.0\,\text{m}$:         $W_{\text{total}} = W_{0 \rightarrow 10} + W_{10 \rightarrow 15} = 2800\,\text{J} + (-700\,\text{J}) = 2100\,\text{J}$

• Work Calculation on a Linear Transition Curve:
  • Problem Statement: Compute the work done by a force $F_x(x)$ as an object translates from $x = 2\,\text{m}$ to $x = 8\,\text{m}$.
  • Graph Analysis:
    • At $x = 2\,\text{m}$, $F_x = +20\,\text{N}$.
    • The force decreases linearly, passing through zero at $x = 6\,\text{m}$ ($F_x(6) = 0\,\text{N}$), and reaches $F_x = -10\,\text{N}$ at $x = 8\,\text{m}$.
  • Positive Area Contribution ($x = 2\,\text{m}$ to $x = 6\,\text{m}$):
    • Right triangle with base $b = 6 - 2 = 4\,\text{m}$ and height $h = 20\,\text{N}$:             $W_1 = \frac{1}{2} (4\,\text{m})(20\,\text{N}) = +40\,\text{J}$
  • Negative Area Contribution ($x = 6\,\text{m}$ to $x = 8\,\text{m}$):
    • Right triangle with base $b = 8 - 6 = 2\,\text{m}$ and height $h = -10\,\text{N}$:             $W_2 = \frac{1}{2} (2\,\text{m})(-10\,\text{N}) = -10\,\text{J}$
  • Net Work:         $W_{\text{net}} = W_1 + W_2 = +40\,\text{J} - 10\,\text{J} = +30\,\text{J}$

Spring Dynamics and Elastic Potential Energy

• Determining the Spring Force Constant via Static Equilibrium:

  • Problem Statement: A vertical spring stretches by an elongation $d = 2.0\,\text{cm} = 2.0 \times 10^{-2}\,\text{m}$ when supporting a suspended object of mass $m = 0.55\,\text{kg}$. Determine the force constant of the spring and the work done by the spring.
  • Equilibrium Condition:
    • Applying the particle in equilibrium model ($a = 0$):             $\sum F_y = F_s - m g = 0 \implies F_s = m g$
    • By Hooke's Law magnitude: $|F_s| = k d$.             $k d = m g$
  • Force Constant Calculation:         $k = \frac{m g}{d} = \frac{(0.55\,\text{kg})(9.8\,\text{m/s}^2)}{2.0 \times 10^{-2}\,\text{m}} = 2.695 \times 10^2\,\text{N/m} \approx 2.7 \times 10^2\,\text{N/m}$
  • Work Done by the Spring During Elongation:
    • Initial position $x_i = 0$, final elongation $x_f = d = 0.02\,\text{m}$.         $W_s = -\Delta U = U_i - U_f = \frac{1}{2} k x_i^2 - \frac{1}{2} k x_f^2$$W_s = 0 - \frac{1}{2} (2.7 \times 10^2\,\text{N/m})(0.02\,\text{m})^2 = -5.4 \times 10^{-2}\,\text{J}$

• Work Done by a Relaxing Ideal Spring:

  • Problem Statement: An ideal spring with force constant $k = 60\,\text{N/m}$ exerts a restoring force $F_x = -k x$. Find the work done by the spring force as an attached object moves from an initial compressed state of $x_i = -0.20\,\text{m}$ back to equilibrium $x_f = 0$.
  • Work Integration:         $W_s = \int_{x_i}^{x_f} (-k x)\,dx = \frac{1}{2} k x_i^2 - \frac{1}{2} k x_f^2$$W_s = \frac{1}{2} (60\,\text{N/m})(-0.20\,\text{m})^2 - \frac{1}{2}(60\,\text{N/m})(0)^2$$W_s = \frac{1}{2}(60)(0.04) = +1.2\,\text{J}$

• Horizontal Block-Spring System Kinematics:
  • Problem Statement: A block of mass $m = 1.6\,\text{kg}$ is connected to a horizontal spring with stiffness $k = 10^3\,\text{N/m}$. The spring is compressed by $x_{\max} = x_i = -2.0\,\text{cm} = -2.0 \times 10^{-2}\,\text{m}$ and released from rest ($v_i = 0$). Calculate the block speed as it passes through the equilibrium point $x_f = 0$ on a frictionless surface.
  • Work Done by the Spring:         $W_s = \frac{1}{2} k x_{\max}^2 = \frac{1}{2} (10^3\,\text{N/m})(-2.0 \times 10^{-2}\,\text{m})^2 = 0.2\,\text{J}$
  • Application of the Work-Kinetic Energy Theorem:         $W_s = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$$0.2\,\text{J} = \frac{1}{2}(1.6\,\text{kg}) v_f^2 - 0$
  • Velocity Calculation:         $v_f = \sqrt{\frac{2 W_s}{m}} = \sqrt{\frac{2(0.2\,\text{J})}{1.6\,\text{kg}}} = \sqrt{\frac{0.4}{1.6}} = \sqrt{0.25} = 0.5\,\text{m/s}$
    • (Note: A typographical error in calculation records the expression as $\sqrt{\frac{2}{1.6}(0.2)^2} = 0.05\,\text{m/s}$; following strict dynamic principles yields $v_f = 0.5\,\text{m/s}$).

Advanced Applications of the Work-Kinetic Energy Theorem

• Two-Dimensional Vector Work-Energy Theorem:
  • Problem Statement: A particle of mass $m = 2.0\,\text{kg}$ experiences a resultant force that changes its velocity from an initial vector $\vec{v}_i = (5\hat{i} - 4\hat{j})\,\text{m/s}$ to a subsequent vector $\vec{v}_f = (7\hat{i} + 3\hat{j})\,\text{m/s}$. Assuming no non-conservative losses, determine the work performed by the resultant force.
  • Kinetic Energy Evaluation via Dot Products:
    • Squared initial speed magnitude:             $v_i^2 = |\vec{v}_i|^2 = (5)^2 + (-4)^2 = 25 + 16 = 41\,\text{m}^2/\text{s}^2$
    • Squared final speed magnitude:             $v_f^2 = |\vec{v}_f|^2 = (7)^2 + (3)^2 = 49 + 9 = 58\,\text{m}^2/\text{s}^2$
  • Net Work:         $W_{\text{net}} = \Delta K = \frac{1}{2} m (v_f^2 - v_i^2)$$W_{\text{net}} = \frac{1}{2}(2.0\,\text{kg})(58 - 41)\,\text{m}^2/\text{s}^2 = (1.0)(17) = 17\,\text{J}$

• Energy Balance on an Incline with External and Frictional Forces:
  • Problem Statement: A block of mass $m = 4.0\,\text{kg}$ is lowered down a $\theta = 37^\circ$ incline over a distance $d = 5.0\,\text{m}$ from point A to point B. A constant horizontal pushing force $F = 10\,\text{N}$ is applied directly toward the incline. Kinetic energy at point A is $K_A = 10\,\text{J}$, and at point B it is $K_B = 20\,\text{J}$. Compute the work done on the block by the force of friction between A and B.
  • Work-Energy Formulation:         $W_{\text{net}} = \Delta K = K_B - K_A = 20\,\text{J} - 10\,\text{J} = 10\,\text{J}$
  • Decomposition of Work Components:         $W_{\text{net}} = W_g + W_F + W_N + W_f$
  • Work Done by Gravity ($W_g$):
    • The vertical downward drop is $h = d \sin(37^\circ) = (5.0\,\text{m})\sin(37^\circ) \approx 5.0(0.6018) = 3.009\,\text{m}$.         $W_g = m g d \sin(37^\circ) = (4.0\,\text{kg})(9.8\,\text{m/s}^2)(5.0\,\text{m})\sin(37^\circ) \approx 117.95\,\text{J}$
  • Work Done by the Horizontal Force ($W_F$):
    • The displacement vector along the plane has a horizontal forward component of $d_x = d \cos(37^\circ)$, while $F$ opposes this motion horizontally in the negative direction:         $W_F = F d \cos(180^\circ - 37^\circ) = -F d \cos(37^\circ)$$W_F = -(10\,\text{N})(5.0\,\text{m})\cos(37^\circ) \approx -50(0.7986) = -39.93\,\text{J}$
  • Work Done by Normal Force ($W_N$):
    • $W_N = 0\,\text{J}$ because the normal force acts perpendicular ($90^\circ$) to the displacement.
  • Solving for Frictional Work ($W_f$):         $W_g + W_F + W_f = \Delta K$$117.95\,\text{J} - 39.93\,\text{J} + W_f = 10\,\text{J}$$78.02\,\text{J} + W_f = 10\,\text{J}$$W_f = 10\,\text{J} - 78.02\,\text{J} = -68.02\,\text{J} \approx -68\,\text{J}$

• Kinetic Energy from a Force Profile Graph:
  • Problem Statement: A single force acts on a body of mass $m = 1.6\,\text{kg}$ along the $x$-axis. Given the body has velocity $v = 5.0\,\text{m/s}$ at position $x = 2.0\,\text{m}$, calculate its kinetic energy at position $x = 5.0\,\text{m}$.
  • Initial Kinetic Energy at $x = 2.0\,\text{m}$:         $K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} (1.6\,\text{kg})(5.0\,\text{m/s})^2 = 0.8 \times 25 = 20\,\text{J}$
  • Work Done Between $x = 2.0\,\text{m}$ and $x = 5.0\,\text{m}$:
    • Segment 1 ($x = 2.0\,\text{m}$ to $x = 4.0\,\text{m}$): Triangular region increasing from $F_x = 0\,\text{N}$ to $F_x = 8\,\text{N}$.             $W_1 = \frac{1}{2} \text{base} \times \text{height} = \frac{1}{2} (4.0 - 2.0\,\text{m})(8\,\text{N}) = \frac{1}{2} (2.0)(8) = 8\,\text{J}$
    • Segment 2 ($x = 4.0\,\text{m}$ to $x = 5.0\,\text{m}$): Rectangular plateau at a constant force of $F_x = 8\,\text{N}$.             $W_2 = \text{width} \times \text{height} = (5.0 - 4.0\,\text{m})(8\,\text{N}) = (1.0)(8) = 8\,\text{J}$
    • Total work accumulated:             $W_{\text{net}} = W_1 + W_2 = 8\,\text{J} + 8\,\text{J} = 16\,\text{J}$
  • Final Kinetic Energy at $x = 5.0\,\text{m}$:         $K_f = K_i + W_{\text{net}} = 20\,\text{J} + 16\,\text{J} = 36\,\text{J}$

Power Output in Mechanical Systems: Elevator Motor Example

• System Specifications:

  • Elevator car mass: $m_{\text{car}} = 1000\,\text{kg}$.
  • Combined passenger mass: $m_{\text{passengers}} = 800\,\text{kg}$.
  • Total system mass: $M = 1000\,\text{kg} + 800\,\text{kg} = 1800\,\text{kg} = 1.8 \times 10^3\,\text{kg}$.
  • Constant downward retarding frictional force: $f = 4000\,\text{N} = 4 \times 10^3\,\text{N}$.

• Motor Power at Constant Speed ($a = 0$):

  • Force Balance: Under the particle in equilibrium model, vertical forces sum to zero:         $\sum F_y = T - f - M g = 0$$T = f + M g$
  • Cable Tension Calculation:         $T = (4 \times 10^3\,\text{N}) + (1.8 \times 10^3\,\text{kg})(9.8\,\text{m/s}^2)$$T = 4000\,\text{N} + 17640\,\text{N} = 2.164 \times 10^4\,\text{N} \approx 2.16 \times 10^4\,\text{N}$
  • Power Output at Constant Speed $v = 3.00\,\text{m/s}$:         $P = T v = (2.164 \times 10^4\,\text{N})(3.00\,\text{m/s}) = 6.48 \times 10^4\,\text{W} = 64.8\,\text{kW}$

• Motor Power Under Upward Acceleration:

  • Dynamic Force Balance: For an upward acceleration of $a = 1.00\,\text{m/s}^2$, apply Newton's second law:         $\sum F_y = T - f - M g = M a$$T = M (a + g) + f$
  • Dynamic Tension Calculation:         $T = (1.8 \times 10^3\,\text{kg})(1.00\,\text{m/s}^2 + 9.8\,\text{m/s}^2) + 4 \times 10^3\,\text{N}$$T = (1.8 \times 10^3)(10.8) + 4 \times 10^3 = 19440\,\text{N} + 4000\,\text{N} = 2.344 \times 10^4\,\text{N} \approx 2.34 \times 10^4\,\text{N}$
  • Instantaneous Power Relation:         $P = T v = (2.34 \times 10^4 v)\,\text{W}$         where $v$ represents the instantaneous velocity in meters per second.
  • Comparison at Instantaneous Speed $v = 3.00\,\text{m/s}$:         $P = (2.34 \times 10^4\,\text{N})(3.00\,\text{m/s}) = 7.02 \times 10^4\,\text{W} = 70.2\,\text{kW}$
  • The accelerating motor delivers significantly higher power ($70.2\,\text{kW}$) compared to steady lifting ($64.8\,\text{kW}$) at the identical operating speed of $3.00\,\text{m/s}$, accounting for the ongoing generation of kinetic energy.