L7: Conservation of Mechanical Energy

Summary:

This is Lecture 7 for the topic “Work, Energy and Power”. You can find all resources related with Work, Energy and Power by clicking on this link

1. Total Mechanical Energy ($E$)

Mechanical energy is the macroscopic energy associated with the motion and configuration of a system. The Total Mechanical Energy ($E$) of a system is defined as the sum of its Kinetic Energy ($K$) and Potential Energy ($U$):

$$E = K + U = \frac{1}{2}mv^2 + U(x, y, z)$$

• Scalar Quantity: Mechanical energy is a scalar, measured in Joules (J).
• Frame Dependent: Because kinetic energy depends on velocity, mechanical energy values are frame-of-reference dependent (though energy conservation laws hold in any inertial frame).

2. Derivation of the Conservation Law

To understand when mechanical energy is conserved, we begin with the fundamental Work-Energy Theorem:

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

In any mechanical system, the total net force can be divided into conservative forces (like gravity, spring force, electrostatic force) and non-conservative forces (like friction, air drag, tension from an active motor, or external applied forces).

$$W_{net} = W_c + W_{nc} = \Delta K$$

By definition, the work done by a conservative force is equal to the negative change in potential energy ($W_c = -\Delta U$). Substituting this into our equation:

$$-\Delta U + W_{nc} = \Delta K$$

Rearranging the terms yields:

$$W_{nc} = \Delta K + \Delta U = \Delta E$$

The General Principle:

The work done by non-conservative forces is equal to the total change in mechanical energy of the system.

The Law of Conservation of Mechanical Energy:

If all internal and external forces acting on a system are conservative—meaning no non-conservative forces do work ($W_{nc} = 0$)—then the total mechanical energy remains strictly constant:

$$\Delta E = 0 \implies K_i + U_i = K_f + U_f$$

3. Classic JEE-Mains/NEET Applications

Application A: The Simple Pendulum

A bob of mass $m$ is suspended by a string of length $L$. It is pulled to an angle $\theta_0$ and released from rest. What is its speed when it passes through the lowest point?

• Analysis: Only gravity and string tension act on the bob. Tension is always perpendicular to instantaneous velocity, so it does zero work ($W_T = 0$). Gravity is conservative. Thus, mechanical energy is conserved.
• Calculation:

• Choose reference $U=0$ at the lowest point.
• Initial state (at release angle $\theta_0$): Height above lowest point is $h = L(1 – \cos\theta_0)$. Velocity $v_i = 0$.
• $E_i = K_i + U_i = 0 + mgL(1 – \cos\theta_0)$
• Final state (at lowest point): Height $h = 0$. Velocity $v_f = v$.
• $E_f = K_f + U_f = \frac{1}{2}mv^2 + 0$
• Equating $E_i = E_f$:
  $$\frac{1}{2}mv^2 = mgL(1 – \cos\theta_0) \implies v = \sqrt{2gL(1 – \cos\theta_0)}$$

Application B: Loop-the-Loop (Minimum Velocity at Top)

A small block slides down a frictionless track and enters a vertical circular loop of radius $R$. What is the minimum release height $H$ from rest so that the block successfully completes the loop without leaving the track?

• Condition at the top of the loop: For the block to stay on the track, the normal reaction must be at least zero ($N \ge 0$). At the exact threshold, $N = 0$, meaning gravity provides the necessary centripetal force:
  $$\frac{mv_{top}^2}{R} = mg \implies v_{top}^2 = gR$$
• Applying Energy Conservation between release point and top of loop:

• $E_{release} = 0 + mgH = mgH$
• $E_{top} = \frac{1}{2}mv_{top}^2 + mg(2R)$
• Equating $E_{release} = E_{top}$:
  $$mgH = \frac{1}{2}m(gR) + 2mgR = \frac{5}{2}mgR \implies H = \frac{5}{2}R$$

JEE-Mains/NEET Practice Questions

Question 1: Pendulum and String Obstacle

A simple pendulum of length $L$ with a bob of mass $m$ is released from a horizontal position ($\theta = 90^\circ$). As it swings down, the string hits a small fixed peg located directly below the point of suspension at a distance $\frac{3L}{4}$ from the top. What is the maximum height reached by the bob on the other side?

Solution:

1. When the bob reaches the lowest point, it has dropped a vertical height $L$. By energy conservation, its velocity $v$ satisfies $\frac{1}{2}mv^2 = mgL \implies v^2 = 2gL$.
2. Below the peg, the radius of the circular swing changes to $r = L – \frac{3L}{4} = \frac{L}{4}$.
3. As it swings up around the peg, energy is still conserved because tension does no work. Let it rise to a height $h’$ above the lowest point.
4. $\frac{1}{2}mv^2 = mgh’ \implies mgL = mgh’ \implies h’ = L$.
5. Since the peg is at depth $\frac{3L}{4}$ below the pivot, the total height reached above the lowest point is $L$, which means it rises to the exact same horizontal level it started from! (Answer: $L$ above lowest point).

Question 2: Spring and Incline Combination

A block of mass $m = 1 \text{ kg}$ is released from rest down a smooth inclined plane of inclination $30^\circ$. After sliding down a distance $d = 2 \text{ m}$, it hits an unstretched ideal spring of constant $k = 200 \text{ N/m}$ aligned parallel to the incline. What is the maximum compression of the spring? (Take $g = 10 \text{ m/s}^2$)

Solution:

1. Let the maximum compression of the spring be $x_m$.
2. Total vertical descent from the initial release point to the point of maximum compression is $(d + x_m) \sin 30^\circ$.
3. Apply conservation of mechanical energy between the initial position and the point of maximum compression (where velocity $v = 0$ momentarily):
   $K_i + U_{gravity, i} + U_{spring, i} = K_f + U_{gravity, f} + U_{spring, f}$
   $0 + 0 + 0 = 0 – mg(d + x_m)\sin 30^\circ + \frac{1}{2}kx_m^2$
4. Substitute values:
   $\frac{1}{2}(200)x_m^2 = (1)(10)(2 + x_m)\left(\frac{1}{2}\right)$
   $100x_m^2 = 5(2 + x_m) \implies 100x_m^2 – 5x_m – 10 = 0$
   Divide by $5$: $20x_m^2 – x_m – 2 = 0$
   Solve via quadratic formula: $x_m = \frac{1 \pm \sqrt{1 – 4(20)(-2)}}{40} = \frac{1 \pm \sqrt{161}}{40} \approx \frac{1 + 12.68}{40} \approx 0.34 \text{ m}$ (or $34 \text{ cm}$).

Question 3: Chain Sliding Off a Table

A uniform chain of mass $M$ and length $L$ is placed on a smooth horizontal table such that a fraction $\frac{1}{n}$ of its length hangs vertically over the edge. If the chain is released from rest, find its velocity when the entire chain just leaves the table.

Solution:

1. Let $y = 0$ be the level of the table surface. Initially, the center of mass of the hanging portion $\frac{L}{n}$ is at a depth of $\frac{L}{2n}$ below the table.
2. Initial potential energy of the system:
   $U_i = -\left(\frac{M}{n}\right)g\left(\frac{L}{2n}\right) = -\frac{MgL}{2n^2}$
   (The portion on the table has $U = 0$).
3. Final state: The entire chain hangs vertically. The center of mass of the whole chain is at a depth of $\frac{L}{2}$ below the table.
   $U_f = -Mg\left(\frac{L}{2}\right) = -\frac{MgL}{2}$
4. Apply conservation of mechanical energy ($K_i + U_i = K_f + U_f$):
   $0 – \frac{MgL}{2n^2} = \frac{1}{2}Mv^2 – \frac{MgL}{2}$
   $\frac{1}{2}v^2 = \frac{gL}{2} \left(1 – \frac{1}{n^2}\right) \implies v = \sqrt{gL \left(1 – \frac{1}{n^2}\right)}$

Question 4: Work-Energy Theorem with Friction (Non-Conservative Force)

A block of mass $2 \text{ kg}$ slides down a rough curved track from a height of $5 \text{ m}$ and reaches the bottom with a speed of $8 \text{ m/s}$. How much mechanical energy was lost due to friction? (Take $g = 10 \text{ m/s}^2$)

Solution:

1. Initial mechanical energy at height $h = 5 \text{ m}$ (with $v_i = 0$):
   $E_i = mgh = (2)(10)(5) = 100 \text{ J}$
2. Final mechanical energy at the bottom ($h = 0$, $v = 8 \text{ m/s}$):
   $E_f = \frac{1}{2}mv^2 + 0 = \frac{1}{2}(2)(8^2) = 64 \text{ J}$
3. The work done by non-conservative forces (friction) equals the change in mechanical energy:
   $W_{nc} = E_f – E_i = 64 – 100 = -36 \text{ J}$
4. Therefore, the mechanical energy lost to friction is $36 \text{ J}$.