SMA Physics

Tag: Work-Energy Theorem

  • L3: Kinetic Energy and the Work-Energy Theorem

    Summary:

    This is Lecture 3 of Chapter Work, Energy and Power. You can find the Lecture 1 and Lecture 2 by clicking on their respective links

    1. Kinetic Energy (KE)

    Kinetic energy is the energy possessed by an object strictly due to its motion. It is a scalar quantity, always positive (or zero), and frame-dependent.

    For a particle of mass $m$ moving with velocity $v$, the kinetic energy $K$ is defined as:

    $$K = \frac{1}{2}mv^2$$

    The Momentum Connection:

    For JEE-Mains, the relationship between kinetic energy and linear momentum ($p = mv$) is heavily tested. By multiplying the numerator and denominator by $m$:

    $$K = \frac{(mv)^2}{2m} = \frac{p^2}{2m}$$

    Application tip: If momentum is increased by $x\%$, use this formula to quickly find the percentage increase in KE.

    2. Deriving the Work-Energy Theorem (WET)

    The Work-Energy Theorem is arguably the most powerful problem-solving tool in mechanics. It bridges dynamics (Newton’s Laws) with kinematics without needing to explicitly calculate acceleration or time.

    Let’s derive it using 1D calculus. From Newton’s Second Law, the net force on an object is $F_{net} = ma$. We can rewrite acceleration as $a = v\frac{dv}{dx}$:

    $$F_{net} = m \left(v\frac{dv}{dx}\right)$$

    Rearranging to integrate with respect to displacement $dx$:

    $$F_{net} dx = mv dv$$

    Integrating both sides from an initial state (position $x_i$, velocity $v_i$) to a final state (position $x_f$, velocity $v_f$):

    $$\int_{x_i}^{x_f} F_{net} dx = m \int_{v_i}^{v_f} v dv$$

    The left side is the definition of Net Work Done ($W_{net}$). The right side integrates to:

    $$W_{net} = m \left[ \frac{v^2}{2} \right]_{v_i}^{v_f} = \frac{1}{2}mv_f^2 – \frac{1}{2}mv_i^2$$

    The Theorem:

    $$W_{net} = \Delta K$$

    The net work done on an object by ALL forces equals the change in its kinetic energy.

    3. Critical Nuances for JEE-Mains

    To avoid common pitfalls in complex problems, drill these rules:

    1. “ALL Forces” means everything: $W_{net}$ includes conservative forces (gravity, spring), non-conservative forces (friction, air drag), external applied forces, and pseudo forces (if working from a non-inertial frame).
      $$W_c + W_{nc} + W_{ext} + W_{pseudo} = \Delta K$$
    2. No Time Required: WET relates velocity and position directly. If a problem asks for velocity as a function of time, you likely need Newton’s laws or impulse-momentum, not WET.
    3. Internal Forces: Work done by internal forces of a system can change its kinetic energy (e.g., an exploding bomb gains KE from internal chemical forces doing work).

    4. Application Strategy: Block on a Rough Incline

    Let’s apply WET to a classic scenario: a block released from rest at the top of a rough inclined plane of length $L$ and angle $\theta$. What is its velocity at the bottom?

    Step 1: Identify all forces doing work.

    • Gravity ($W$ in the diagram): Has a component along the displacement. Does positive work.
    • Normal Force ($N$): Perpendicular to displacement. Does zero work.
    • Kinetic Friction ($f$): Opposes displacement. Does negative work.

    Step 2: Calculate work done by each.

    • $W_{gravity} = (mg \sin \theta) \cdot L$
    • $W_{friction} = -f \cdot L = -(\mu mg \cos \theta) \cdot L$

    Step 3: Apply $W_{net} = \Delta K$.

    $$mgL \sin \theta – \mu mgL \cos \theta = \frac{1}{2}mv_f^2 – 0$$

    $$v_f = \sqrt{2gL(\sin \theta – \mu \cos \theta)}$$

    JEE-Mains Practice Questions

    Question 1: The Momentum-KE Relation

    If the linear momentum of a body is increased by 50%, what will be the percentage increase in its kinetic energy?

    Solution:

    Let initial momentum be $p_1 = p$. Initial KE is $K_1 = \frac{p^2}{2m}$.

    New momentum $p_2 = p + 0.50p = 1.5p$.

    New KE $K_2 = \frac{(1.5p)^2}{2m} = 2.25 \left(\frac{p^2}{2m}\right) = 2.25 K_1$.

    Percentage increase = $\left(\frac{K_2 – K_1}{K_1}\right) \times 100$

    Percentage increase = $(2.25 – 1) \times 100 = 125\%$

    Question 2: Penetration (Frictional Work)

    A bullet of mass $20$ g traveling at $400$ m/s strikes a massive wooden block. If it penetrates $20$ cm into the block before coming to rest, what is the average resistive force exerted by the block?

    Solution:

    Apply WET: $W_{net} = \Delta K$

    The only force doing work is the resistive force $F_r$, acting opposite to displacement $d$.

    $-F_r \cdot d = K_f – K_i$

    $-F_r (0.2) = 0 – \frac{1}{2}(0.020)(400)^2$

    $-F_r (0.2) = – \frac{1}{2}(0.020)(160000)$

    $-F_r (0.2) = -1600$

    $F_r = 8000$ N

    Question 3: Variable Force Application

    A particle of mass $0.5$ kg is subjected to a force which varies with distance $x$ as $F = (3x^2 + 4)$ N. If the particle starts from rest at $x = 0$, what is its kinetic energy when it reaches $x = 2$ m?

    Solution:

    Since force is variable, use the integral form: $W = \int F dx$.

    $W = \int_{0}^{2} (3x^2 + 4) dx$

    $W = \left[ x^3 + 4x \right]_{0}^{2}$

    $W = (2^3 + 4(2)) – 0 = 8 + 8 = 16$ J

    By WET, $W = \Delta K \implies 16 = K_f – 0$.

    Final Kinetic Energy = $16$ J