## Work-Energy Principle in a Resistive Medium ## Introduction The standard Work-Energy Principle states that the net work done on an object equals its change in kinetic energy. However, in most real-world scenarios, objects move through a **resistive medium**, where forces like **friction** or **air resistance** oppose the motion. These non-conservative forces do negative work, dissipating mechanical energy from the system, usually in the form of heat. This section extends the Work-Energy Principle to account for the work done by such resistive forces. ## Net Work in the Presence of Resistive Forces The **net work ($W_{net}$)** is the algebraic sum of the work done by all forces acting on an object. This includes the work done by applied forces ($W_{applied}$) and the work done by resistive forces ($W_{resistive}$). $$W_{net} = W_{applied} + W_{resistive}$$ The Work-Energy Principle remains fundamentally the same: $$W_{net} = \Delta K.E.$$ ## Work Done by Resistive Forces Resistive forces, such as friction or air drag, always act in the direction opposite to the object's displacement. This means the angle ($\theta$) between the resistive force vector and the displacement vector is $180^\circ$. Since $\cos(180^\circ) = -1$, the work done by a resistive force is always **negative**. $$W_{resistive} = F_{resistive} \cdot d \cdot \cos(180^\circ) = -F_{resistive} \cdot d$$ This negative work represents a removal of mechanical energy from the object. ## The Extended Work-Energy Equation Combining these ideas, the Work-Energy Principle in a resistive medium can be written as: $$W_{applied} + W_{resistive} = \Delta K.E.$$ $$W_{applied} - (F_{resistive} \cdot d) = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$ For a body falling through a height $h$ in a resistive medium (like air), the loss in potential energy is used in doing work against friction ($fh$) and increasing kinetic energy: $$mgh = \frac{1}{2}mv^2 + fh$$ This equation shows that the work done by the applied force (or gravity) does not fully translate into kinetic energy; some of it is "lost" to the work done against the resistive force. ## Connection to Conservation of Energy The work done by non-conservative forces like friction is equal to the change in the total mechanical energy of the system. $$W_{resistive} = \Delta E_{mech} = \Delta K.E. + \Delta P.E.$$ Since $W_{resistive}$ is negative, the total mechanical energy of the system decreases. This "lost" energy is converted into other forms, primarily thermal energy (heat). ## Example Problem A box of mass $m = 10$ kg is pushed across a rough horizontal surface. An applied force of $F_{applied} = 50$ N moves the box from rest over a distance of $d = 5$ m. The constant frictional force is $F_{friction} = 20$ N. Find the final speed of the box. **Solution:** 1. **Calculate the work done by the applied force ($W_{applied}$)**: This force is in the direction of motion ($\theta = 0^\circ$). $$W_{applied} = F_{applied} \cdot d = (50 \text{ N}) \cdot (5 \text{ m}) = 250 \text{ J}$$ 2. **Calculate the work done by the frictional force ($W_{friction}$)**: This force opposes the motion ($\theta = 180^\circ$). $$W_{friction} = F_{friction} \cdot d \cdot \cos(180^\circ) = (20 \text{ N}) \cdot (5 \text{ m}) \cdot (-1) = -100 \text{ J}$$ 3. **Calculate the net work done ($W_{net}$)**: $$W_{net} = W_{applied} + W_{friction} = 250 \text{ J} + (-100 \text{ J}) = 150 \text{ J}$$ 4. **Apply the Work-Energy Principle ($W_{net} = \Delta K.E.$)**: The change in kinetic energy is equal to the net work. The initial kinetic energy is zero since the box starts from rest ($v_i = 0$). $$\Delta K.E. = K.E._f - K.E._i = \frac{1}{2}mv_f^2 - 0 = 150 \text{ J}$$ 5. **Solve for the final velocity ($v_f$)**: $$\frac{1}{2}(10 \text{ kg})v_f^2 = 150 \text{ J}$$ $$5v_f^2 = 150$$ $$v_f^2 = 30 \implies v_f = \sqrt{30} \approx 5.48 \text{ m/s}$$ ## Possible Questions and Answers **Q:** What happens to the energy that is "lost" due to friction? **A:** It is not truly lost; it is transformed into other forms of energy, primarily thermal energy (heat). This is why the brakes on a car or the tires during a skid get very hot. **Q:** If an object is pushed at a constant velocity across a rough surface, what is the net work done on it? **A:** The net work is zero. Since the velocity is constant, the change in kinetic energy is zero. This means the positive work done by the applied force is exactly cancelled out by the negative work done by the force of friction. | **Equation** | **Description** | | :--- | :--- | | **$W_{net} = W_{applied} + W_{resistive}$** | The net work is the sum of work from all forces. | | **$W_{net} = \Delta K.E.$** | The net work equals the change in kinetic energy. | | **$W_{resistive} = -F_{resistive} \cdot d$** | Work done by a resistive force is always negative. | | **$mgh = \frac{1}{2}mv^2 + fh$** | Energy conservation for a falling body in air friction. |

Work Energy Theorem In Resistivity Medium | Work And Kinetic Energy | Physics 11